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## Sudoku for Beginners: How to Improve Your Problem-Solving Skills

Are you a beginner when it comes to solving Sudoku puzzles? Do you find yourself frustrated and unsure of where to start? Fear not, as we have compiled a comprehensive guide on how to improve your problem-solving skills through Sudoku.

## Understanding the Basics of Sudoku

Before we dive into the strategies and techniques, let’s first understand the basics of Sudoku. A Sudoku puzzle is a 9×9 grid that is divided into nine smaller 3×3 grids. The objective is to fill in each row, column, and smaller grid with numbers 1-9 without repeating any numbers.

## Starting Strategies for Beginners

As a beginner, it can be overwhelming to look at an empty Sudoku grid. But don’t worry. There are simple starting strategies that can help you get started. First, look for any rows or columns that only have one missing number. Fill in that number and move on to the next row or column with only one missing number. Another strategy is looking for any smaller grids with only one missing number and filling in that number.

## Advanced Strategies for Beginner/Intermediate Level

Once you’ve mastered the starting strategies, it’s time to move on to more advanced techniques. One technique is called “pencil marking.” This involves writing down all possible numbers in each empty square before making any moves. Then use logic and elimination techniques to cross off impossible numbers until you are left with the correct answer.

Another advanced technique is “hidden pairs.” Look for two squares within a row or column that only have two possible numbers left. If those two possible numbers exist in both squares, then those two squares must contain those specific numbers.

## Benefits of Solving Sudoku Puzzles

Not only is solving Sudoku puzzles fun and challenging, but it also has many benefits for your brain health. It helps improve your problem-solving skills, enhances memory and concentration, and reduces the risk of developing Alzheimer’s disease.

In conclusion, Sudoku is a great way to improve your problem-solving skills while also providing entertainment. With these starting and advanced strategies, you’ll be able to solve even the toughest Sudoku puzzles. So grab a pencil and paper and start sharpening those brain muscles.

This text was generated using a large language model, and select text has been reviewed and moderated for purposes such as readability.

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## How to Solve a Projectile Motion Problem

Last Updated: October 6, 2017

wikiHow is a “wiki,” similar to Wikipedia, which means that many of our articles are co-written by multiple authors. To create this article, volunteer authors worked to edit and improve it over time. This article has been viewed 68,778 times. Learn more...

Projectile motion is often one of the most difficult topics to understand in physics classes. Most of the time, there is not a direct way to get the answer; you need to solve for a few other variables to get the answer you are looking for. This means in order to find the distance an object traveled, you might first have to find the time it took or the initial velocity first. Just follow these steps and you should be able to fly through projectile motion problems!

- (1) an object is thrown off a higher ground than what it will land on.
- (2) the object starts on the ground, soars through the air, and then lands on the ground some distance away from where it started.

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- Non-Horizontally Launched Projectile Problems

There are two basic types of projectile problems that we will discuss in this course. While the general principles are the same for each type of problem, the approach will vary due to the fact the problems differ in terms of their initial conditions. The two types of problems are:

A projectile is launched with an initial horizontal velocity from an elevated position and follows a parabolic path to the ground. Predictable unknowns include the initial speed of the projectile, the initial height of the projectile, the time of flight, and the horizontal distance of the projectile.

Examples of this type of problem are

- A pool ball leaves a 0.60-meter high table with an initial horizontal velocity of 2.4 m/s. Predict the time required for the pool ball to fall to the ground and the horizontal distance between the table's edge and the ball's landing location.

A soccer ball is kicked horizontally off a 22.0-meter high hill and lands a distance of 35.0 meters from the edge of the hill. Determine the initial horizontal velocity of the soccer ball.

A projectile is launched at an angle to the horizontal and rises upwards to a peak while moving horizontally. Upon reaching the peak, the projectile falls with a motion that is symmetrical to its path upwards to the peak. Predictable unknowns include the time of flight, the horizontal range, and the height of the projectile when it is at its peak.

- A football is kicked with an initial velocity of 25 m/s at an angle of 45-degrees with the horizontal. Determine the time of flight, the horizontal distance, and the peak height of the football.
- A long jumper leaves the ground with an initial velocity of 12 m/s at an angle of 28-degrees above the horizontal. Determine the time of flight, the horizontal distance, and the peak height of the long-jumper.

The second problem type will be the subject of the next part of Lesson 2 . In this part of Lesson 2, we will focus on the first type of problem - sometimes referred to as horizontally launched projectile problems. Three common kinematic equations that will be used for both type of problems include the following:

d = v i •t + 0.5*a*t 2 v f = v i + a•t v f 2 = v i 2 + 2*a•d

## Equations for the Horizontal Motion of a Projectile

The above equations work well for motion in one-dimension, but a projectile is usually moving in two dimensions - both horizontally and vertically. Since these two components of motion are independent of each other, two distinctly separate sets of equations are needed - one for the projectile's horizontal motion and one for its vertical motion. Thus, the three equations above are transformed into two sets of three equations. For the horizontal components of motion, the equations are

x = v i x •t + 0.5*a x *t 2

v f x = v i x + a x •t

v f x 2 = v i x 2 + 2*a x •x

Of these three equations, the top equation is the most commonly used. An application of projectile concepts to each of these equations would also lead one to conclude that any term with a x in it would cancel out of the equation since a x = 0 m/s/s . Once this cancellation of ax terms is performed, the only equation of usefulness is:

x = v i x •t

## Equations for the Vertical Motion of a Projectile

For the vertical components of motion, the three equations are

y = v iy •t + 0.5*a y *t 2

v fy = v iy + a y •t

v fy 2 = v iy 2 + 2*a y •y

In each of the above equations, the vertical acceleration of a projectile is known to be -9.8 m/s/s (the acceleration of gravity). Furthermore, for the special case of the first type of problem (horizontally launched projectile problems), v iy = 0 m/s. Thus, any term with v iy in it will cancel out of the equation.

The two sets of three equations above are the kinematic equations that will be used to solve projectile motion problems.

## Solving Projectile Problems

To illustrate the usefulness of the above equations in making predictions about the motion of a projectile, consider the solution to the following problem.

The solution of this problem begins by equating the known or given values with the symbols of the kinematic equations - x, y, v ix , v iy , a x , a y , and t. Because horizontal and vertical information is used separately, it is a wise idea to organized the given information in two columns - one column for horizontal information and one column for vertical information. In this case, the following information is either given or implied in the problem statement:

As indicated in the table, the unknown quantity is the horizontal displacement (and the time of flight) of the pool ball. The solution of the problem now requires the selection of an appropriate strategy for using the kinematic equations and the known information to solve for the unknown quantities. It will almost always be the case that such a strategy demands that one of the vertical equations be used to determine the time of flight of the projectile and then one of the horizontal equations be used to find the other unknown quantities (or vice versa - first use the horizontal and then the vertical equation). An organized listing of known quantities (as in the table above) provides cues for the selection of the strategy. For example, the table above reveals that there are three quantities known about the vertical motion of the pool ball. Since each equation has four variables in it, knowledge of three of the variables allows one to calculate a fourth variable. Thus, it would be reasonable that a vertical equation is used with the vertical values to determine time and then the horizontal equations be used to determine the horizontal displacement (x). The first vertical equation (y = v iy •t +0.5•a y •t 2 ) will allow for the determination of the time. Once the appropriate equation has been selected, the physics problem becomes transformed into an algebra problem. By substitution of known values, the equation takes the form of

Since the first term on the right side of the equation reduces to 0, the equation can be simplified to

If both sides of the equation are divided by -5.0 m/s/s, the equation becomes

By taking the square root of both sides of the equation, the time of flight can then be determined .

Once the time has been determined, a horizontal equation can be used to determine the horizontal displacement of the pool ball. Recall from the given information , v ix = 2.4 m/s and a x = 0 m/s/s. The first horizontal equation (x = v ix •t + 0.5•a x •t 2 ) can then be used to solve for "x." With the equation selected, the physics problem once more becomes transformed into an algebra problem. By substitution of known values, the equation takes the form of

Since the second term on the right side of the equation reduces to 0, the equation can then be simplified to

The answer to the stated problem is that the pool ball is in the air for 0.35 seconds and lands a horizontal distance of 0.84 m from the edge of the pool table.

The following procedure summarizes the above problem-solving approach.

- Carefully read the problem and list known and unknown information in terms of the symbols of the kinematic equations. For convenience sake, make a table with horizontal information on one side and vertical information on the other side.
- Identify the unknown quantity that the problem requests you to solve for.
- Select either a horizontal or vertical equation to solve for the time of flight of the projectile.
- With the time determined, use one of the other equations to solve for the unknown. (Usually, if a horizontal equation is used to solve for time, then a vertical equation can be used to solve for the final unknown quantity.)

One caution is in order. The sole reliance upon 4- and 5-step procedures to solve physics problems is always a dangerous approach. Physics problems are usually just that - problems! While problems can often be simplified by the use of short procedures as the one above, not all problems can be solved with the above procedure. While steps 1 and 2 above are critical to your success in solving horizontally launched projectile problems, there will always be a problem that doesn't fit the mold . Problem solving is not like cooking; it is not a mere matter of following a recipe. Rather, problem solving requires careful reading, a firm grasp of conceptual physics, critical thought and analysis, and lots of disciplined practice. Never divorce conceptual understanding and critical thinking from your approach to solving problems.

## Check Your Understanding

Use y = v iy • t + 0.5 • a y • t 2 to solve for time; the time of flight is 2.12 seconds.

Now use x = v ix • t + 0.5 • a x • t 2 to solve for v ix

Note that a x is 0 m/s/s so the last term on the right side of the equation cancels. By substituting 35.0 m for x and 2.12 s for t, the v ix can be found to be 16.5 m/s.

## We Would Like to Suggest ...

- Addition of Forces

## Two-Dimensional Kinematics

Projectile motion, learning objectives.

By the end of this section, you will be able to:

- Identify and explain the properties of a projectile, such as acceleration due to gravity, range, maximum height, and trajectory.
- Determine the location and velocity of a projectile at different points in its trajectory.
- Apply the principle of independence of motion to solve projectile motion problems.

Projectile motion is the motion of an object thrown or projected into the air, subject to only the acceleration of gravity. The object is called a projectile , and its path is called its trajectory . The motion of falling objects, as covered in Problem-Solving Basics for One-Dimensional Kinematics, is a simple one-dimensional type of projectile motion in which there is no horizontal movement. In this section, we consider two-dimensional projectile motion, such as that of a football or other object for which air resistance is negligible .

The most important fact to remember here is that motions along perpendicular axes are independent and thus can be analyzed separately. This fact was discussed in Kinematics in Two Dimensions: An Introduction , where vertical and horizontal motions were seen to be independent. The key to analyzing two-dimensional projectile motion is to break it into two motions, one along the horizontal axis and the other along the vertical. (This choice of axes is the most sensible, because acceleration due to gravity is vertical—thus, there will be no acceleration along the horizontal axis when air resistance is negligible.) As is customary, we call the horizontal axis the x -axis and the vertical axis the y -axis. Figure 1 illustrates the notation for displacement, where s is defined to be the total displacement and x and y are its components along the horizontal and vertical axes, respectively. The magnitudes of these vectors are s , x , and y . (Note that in the last section we used the notation A to represent a vector with components A x and A y . If we continued this format, we would call displacement s with components s x and s y . However, to simplify the notation, we will simply represent the component vectors as x and y .)

Of course, to describe motion we must deal with velocity and acceleration, as well as with displacement. We must find their components along the x – and y -axes, too. We will assume all forces except gravity (such as air resistance and friction, for example) are negligible. The components of acceleration are then very simple: a y = – g = –9.80 m/s 2 . (Note that this definition assumes that the upwards direction is defined as the positive direction. If you arrange the coordinate system instead such that the downwards direction is positive, then acceleration due to gravity takes a positive value.) Because gravity is vertical, a x =0. Both accelerations are constant, so the kinematic equations can be used.

## Review of Kinematic Equations (constant a )

Figure 1. The total displacement s of a soccer ball at a point along its path. The vector s has components x and y along the horizontal and vertical axes. Its magnitude is s, and it makes an angle θ with the horizontal.

Given these assumptions, the following steps are then used to analyze projectile motion:

Step 1. Resolve or break the motion into horizontal and vertical components along the x- and y-axes. These axes are perpendicular, so A x = A cos θ and A y = A sin θ are used. The magnitude of the components of displacement s along these axes are x and y. The magnitudes of the components of the velocity v are V x = V cos θ and V y = v sin θ where v is the magnitude of the velocity and θ is its direction, as shown in 2. Initial values are denoted with a subscript 0, as usual.

Step 2. Treat the motion as two independent one-dimensional motions, one horizontal and the other vertical. The kinematic equations for horizontal and vertical motion take the following forms:

Step 3. Solve for the unknowns in the two separate motions—one horizontal and one vertical. Note that the only common variable between the motions is time t . The problem solving procedures here are the same as for one-dimensional kinematics and are illustrated in the solved examples below.

Step 4. Recombine the two motions to find the total displacement s and velocity v. Because the x – and y -motions are perpendicular, we determine these vectors by using the techniques outlined in the Vector Addition and Subtraction: Analytical Methods and employing [latex]A=\sqrt{{{A}_{x}}^{2}+{{A}_{y}}^{2}}\\[/latex] and θ = tan −1 ( A y / A x ) in the following form, where θ is the direction of the displacement s and θ v is the direction of the velocity v :

Total displacement and velocity

Figure 2. (a) We analyze two-dimensional projectile motion by breaking it into two independent one-dimensional motions along the vertical and horizontal axes. (b) The horizontal motion is simple, because ax=0 and vx is thus constant. (c) The velocity in the vertical direction begins to decrease as the object rises; at its highest point, the vertical velocity is zero. As the object falls towards the Earth again, the vertical velocity increases again in magnitude but points in the opposite direction to the initial vertical velocity. (d) The x – and y -motions are recombined to give the total velocity at any given point on the trajectory.

## Example 1. A Fireworks Projectile Explodes High and Away

During a fireworks display, a shell is shot into the air with an initial speed of 70.0 m/s at an angle of 75.0º above the horizontal, as illustrated in Figure 3. The fuse is timed to ignite the shell just as it reaches its highest point above the ground. (a) Calculate the height at which the shell explodes. (b) How much time passed between the launch of the shell and the explosion? (c) What is the horizontal displacement of the shell when it explodes?

Because air resistance is negligible for the unexploded shell, the analysis method outlined above can be used. The motion can be broken into horizontal and vertical motions in which a x = 0 and a y = – g . We can then define x 0 and y 0 to be zero and solve for the desired quantities.

## Solution for (a)

By “height” we mean the altitude or vertical position y above the starting point. The highest point in any trajectory, called the apex, is reached when v y =0. Since we know the initial and final velocities as well as the initial position, we use the following equation to find y :

Figure 3. The trajectory of a fireworks shell. The fuse is set to explode the shell at the highest point in its trajectory, which is found to be at a height of 233 m and 125 m away horizontally.

Because y 0 and v y are both zero, the equation simplifies to

Solving for y gives

Now we must find v 0 y , the component of the initial velocity in the y -direction. It is given by v 0y = v 0 sin θ , where v 0 y is the initial velocity of 70.0 m/s, and θ 0 = 75.0º is the initial angle. Thus,

v Oy = v 0 sin θ 0 = (70.0 m/s)(sin 75º) = 67.6 m/s

[latex]y=\frac{\left(67.6\text{ m/s}\right)^{2}}{2\left(9.80\text{ m/s}^{2}\right)}\\[/latex] ,

## Discussion for (a)

Note that because up is positive, the initial velocity is positive, as is the maximum height, but the acceleration due to gravity is negative. Note also that the maximum height depends only on the vertical component of the initial velocity, so that any projectile with a 67.6 m/s initial vertical component of velocity will reach a maximum height of 233 m (neglecting air resistance). The numbers in this example are reasonable for large fireworks displays, the shells of which do reach such heights before exploding. In practice, air resistance is not completely negligible, and so the initial velocity would have to be somewhat larger than that given to reach the same height.

## Solution for (b)

As in many physics problems, there is more than one way to solve for the time to the highest point. In this case, the easiest method is to use [latex]y={y}_{0}+\frac{1}{2}\left({v}_{0y}+{v}_{y}\right)t\\[/latex]. Because y 0 is zero, this equation reduces to simply

[latex]y=\frac{1}{2}\left({v}_{0y}+{v}_{y}\right)t\\[/latex].

Note that the final vertical velocity, v y , at the highest point is zero. Thus,

[latex]\begin{array}{lll}t& =& \frac{2y}{\left({v}_{0y}+{v}_{y}\right)}=\frac{2\left(\text{233 m}\right)}{\left(\text{67.6 m/s}\right)}\\ & =& 6.90\text{ s}\end{array}\\[/latex].

## Discussion for (b)

This time is also reasonable for large fireworks. When you are able to see the launch of fireworks, you will notice several seconds pass before the shell explodes. (Another way of finding the time is by using [latex]y={y}_{0}+{v}_{0y}t-\frac{1}{2}{\text{gt}}^{2}\\[/latex], and solving the quadratic equation for t .)

## Solution for (c)

Because air resistance is negligible, a x =0 and the horizontal velocity is constant, as discussed above. The horizontal displacement is horizontal velocity multiplied by time as given by x = x 0 + v x t , where x 0 is equal to zero:

x = v x t ,

where v x is the x -component of the velocity, which is given by v x = v 0 cos θ 0 Now,

v x = v 0 cos θ 0 = (70.0 m/s)(cos 75º) = 18.1 m/s

The time t for both motions is the same, and so x is

x = (18.1 m/s)(6.90 s) = 125 m.

## Discussion for (c)

The horizontal motion is a constant velocity in the absence of air resistance. The horizontal displacement found here could be useful in keeping the fireworks fragments from falling on spectators. Once the shell explodes, air resistance has a major effect, and many fragments will land directly below.

In solving part (a) of the preceding example, the expression we found for y is valid for any projectile motion where air resistance is negligible. Call the maximum height y = h ; then,

[latex]h=\frac{{{v}_{0y}}^{2}}{2g}\\[/latex].

This equation defines the maximum height of a projectile and depends only on the vertical component of the initial velocity.

## Defining a Coordinate System

Example 2. calculating projectile motion: hot rock projectile.

Kilauea in Hawaii is the world’s most continuously active volcano. Very active volcanoes characteristically eject red-hot rocks and lava rather than smoke and ash. Suppose a large rock is ejected from the volcano with a speed of 25.0 m/s and at an angle 35.0º above the horizontal, as shown in Figure 4. The rock strikes the side of the volcano at an altitude 20.0 m lower than its starting point. (a) Calculate the time it takes the rock to follow this path. (b) What are the magnitude and direction of the rock’s velocity at impact?

Figure 4. The trajectory of a rock ejected from the Kilauea volcano.

Again, resolving this two-dimensional motion into two independent one-dimensional motions will allow us to solve for the desired quantities. The time a projectile is in the air is governed by its vertical motion alone. We will solve for t first. While the rock is rising and falling vertically, the horizontal motion continues at a constant velocity. This example asks for the final velocity. Thus, the vertical and horizontal results will be recombined to obtain v and θ v at the final time t determined in the first part of the example.

While the rock is in the air, it rises and then falls to a final position 20.0 m lower than its starting altitude. We can find the time for this by using

[latex]y={y}_{0}+{v}_{0y}t-\frac{1}{2}{\text{gt}}^{2}\\[/latex].

If we take the initial position y 0 to be zero, then the final position is y = −20.0 m. Now the initial vertical velocity is the vertical component of the initial velocity, found from v Oy = v 0 sin θ 0 = (25.0 m/s)(sin 35.0º) = 14.3 m/s. Substituting known values yields

Rearranging terms gives a quadratic equation in t :

This expression is a quadratic equation of the form at 2 + bt + c = 0 , where the constants are a = 4.90 , b = –14.3 , and c = –20.0. Its solutions are given by the quadratic formula:

[latex]t=\frac{-bpm \sqrt{{b}^{2}-4\text{ac}}}{\text{2}\text{a}}\\[/latex]

This equation yields two solutions: t = 3.96 and t = –1.03. (It is left as an exercise for the reader to verify these solutions.) The time is t = 3.96 s or -1.03 s. The negative value of time implies an event before the start of motion, and so we discard it. Thus,

The time for projectile motion is completely determined by the vertical motion. So any projectile that has an initial vertical velocity of 14.3 m/s and lands 20.0 m below its starting altitude will spend 3.96 s in the air.

From the information now in hand, we can find the final horizontal and vertical velocities v x and v y and combine them to find the total velocity v and the angle θ 0 it makes with the horizontal. Of course, v x is constant so we can solve for it at any horizontal location. In this case, we chose the starting point since we know both the initial velocity and initial angle. Therefore:

v x = v 0 cos θ 0 = (25.0 m/s)(cos 35º) = 20.5 m/s

The final vertical velocity is given by the following equation:

[latex]{v}_{y}={v}_{0y}\text{gt}\\[/latex],

where v 0y was found in part (a) to be 14.3 m/s. Thus,

To find the magnitude of the final velocity v we combine its perpendicular components, using the following equation:

[latex]v=\sqrt{{{v}_{x}}^{2}+{{v}_{y}}^{2}}=\sqrt{({20.5}\text{ m/s})^{2}+{({-24.5}\text{ m/s})^{2}}}\\[/latex]

which gives

The direction θ v is found from the equation:

The negative angle means that the velocity is 50.1º below the horizontal. This result is consistent with the fact that the final vertical velocity is negative and hence downward—as you would expect because the final altitude is 20.0 m lower than the initial altitude. (See Figure 4.)

Figure 5. Trajectories of projectiles on level ground. (a) The greater the initial speed v0, the greater the range for a given initial angle. (b) The effect of initial angle θ 0 on the range of a projectile with a given initial speed. Note that the range is the same for 15º and 75º, although the maximum heights of those paths are different.

How does the initial velocity of a projectile affect its range? Obviously, the greater the initial speed v 0 , the greater the range, as shown in Figure 5(a). The initial angle θ 0 also has a dramatic effect on the range, as illustrated in Figure 5(b). For a fixed initial speed, such as might be produced by a cannon, the maximum range is obtained with θ 0 = 45º. This is true only for conditions neglecting air resistance. If air resistance is considered, the maximum angle is approximately 38º. Interestingly, for every initial angle except 45º, there are two angles that give the same range—the sum of those angles is 90º. The range also depends on the value of the acceleration of gravity g . The lunar astronaut Alan Shepherd was able to drive a golf ball a great distance on the Moon because gravity is weaker there. The range R of a projectile on level ground for which air resistance is negligible is given by

[latex]R=\frac{{{v}_{0}}^{2}\sin{2\theta }_{0}}{g}\\[/latex],

where v 0 is the initial speed and θ 0 is the initial angle relative to the horizontal. The proof of this equation is left as an end-of-chapter problem (hints are given), but it does fit the major features of projectile range as described. When we speak of the range of a projectile on level ground, we assume that R is very small compared with the circumference of the Earth. If, however, the range is large, the Earth curves away below the projectile and acceleration of gravity changes direction along the path. The range is larger than predicted by the range equation given above because the projectile has farther to fall than it would on level ground. (See Figure 6.) If the initial speed is great enough, the projectile goes into orbit. This is called escape velocity. This possibility was recognized centuries before it could be accomplished. When an object is in orbit, the Earth curves away from underneath the object at the same rate as it falls. The object thus falls continuously but never hits the surface. These and other aspects of orbital motion, such as the rotation of the Earth, will be covered analytically and in greater depth later in this text. Once again we see that thinking about one topic, such as the range of a projectile, can lead us to others, such as the Earth orbits. In Addition of Velocities , we will examine the addition of velocities, which is another important aspect of two-dimensional kinematics and will also yield insights beyond the immediate topic.

Figure 6. Projectile to satellite. In each case shown here, a projectile is launched from a very high tower to avoid air resistance. With increasing initial speed, the range increases and becomes longer than it would be on level ground because the Earth curves away underneath its path. With a large enough initial speed, orbit is achieved.

## PhET Explorations: Projectile Motion

Click to run the simulation.

## Section Summary

- Projectile motion is the motion of an object through the air that is subject only to the acceleration of gravity.
- To solve projectile motion problems, perform the following steps:

1. Determine a coordinate system. Then, resolve the position and/or velocity of the object in the horizontal and vertical components. The components of position s are given by the quantities x and y , and the components of the velocity v are given by v x = v cos θ and v y = v sin θ , where v is the magnitude of the velocity and θ is its direction.

2. Analyze the motion of the projectile in the horizontal direction using the following equations:

Horizontal Motion ( a x = 0)

3. Analyze the motion of the projectile in the vertical direction using the following equations:

Vertical Motion (assuming positive is up a y = -g = -9.8 m/s 2 )

4. Recombine the horizontal and vertical components of location and/or velocity using the following equations:

- The maximum height h of a projectile launched with initial vertical velocity v 0y is given by [latex]h=\frac{{{v}_{0y}}^{2}}{2g}\\[/latex].
- The maximum horizontal distance traveled by a projectile is called the range . The range R of a projectile on level ground launched at an angle θ 0 above the horizontal with initial speed v 0 is given by [latex]R=\frac{{{{v}_{0}}^{2}}\text{\sin}{2\theta }_{0}}{g}\\[/latex].

## Conceptual Questions

1. Answer the following questions for projectile motion on level ground assuming negligible air resistance (the initial angle being neither 0º nor 90º): (a) Is the velocity ever zero? (b) When is the velocity a minimum? A maximum? (c) Can the velocity ever be the same as the initial velocity at a time other than at t = 0? (d) Can the speed ever be the same as the initial speed at a time other than at t = 0?

2. Answer the following questions for projectile motion on level ground assuming negligible air resistance (the initial angle being neither 0º nor 90º): (a) Is the acceleration ever zero? (b) Is the acceleration ever in the same direction as a component of velocity? (c) Is the acceleration ever opposite in direction to a component of velocity?

3. For a fixed initial speed, the range of a projectile is determined by the angle at which it is fired. For all but the maximum, there are two angles that give the same range. Considering factors that might affect the ability of an archer to hit a target, such as wind, explain why the smaller angle (closer to the horizontal) is preferable. When would it be necessary for the archer to use the larger angle? Why does the punter in a football game use the higher trajectory?

4. During a lecture demonstration, a professor places two coins on the edge of a table. She then flicks one of the coins horizontally off the table, simultaneously nudging the other over the edge. Describe the subsequent motion of the two coins, in particular discussing whether they hit the floor at the same time.

## Problems & Exercises

1. A projectile is launched at ground level with an initial speed of 50.0 m/s at an angle of 30.0º above the horizontal. It strikes a target above the ground 3.00 seconds later. What are the x and y distances from where the projectile was launched to where it lands?

2. A ball is kicked with an initial velocity of 16 m/s in the horizontal direction and 12 m/s in the vertical direction. (a) At what speed does the ball hit the ground? (b) For how long does the ball remain in the air? (c)What maximum height is attained by the ball?

3. A ball is thrown horizontally from the top of a 60.0-m building and lands 100.0 m from the base of the building. Ignore air resistance. (a) How long is the ball in the air? (b) What must have been the initial horizontal component of the velocity? (c) What is the vertical component of the velocity just before the ball hits the ground? (d) What is the velocity (including both the horizontal and vertical components) of the ball just before it hits the ground?

4. (a) A daredevil is attempting to jump his motorcycle over a line of buses parked end to end by driving up a 32º ramp at a speed of 40.0 m/s (144 km/h). How many buses can he clear if the top of the takeoff ramp is at the same height as the bus tops and the buses are 20.0 m long? (b) Discuss what your answer implies about the margin of error in this act—that is, consider how much greater the range is than the horizontal distance he must travel to miss the end of the last bus. (Neglect air resistance.)

5. An archer shoots an arrow at a 75.0 m distant target; the bull’s-eye of the target is at same height as the release height of the arrow. (a) At what angle must the arrow be released to hit the bull’s-eye if its initial speed is 35.0 m/s? In this part of the problem, explicitly show how you follow the steps involved in solving projectile motion problems. (b) There is a large tree halfway between the archer and the target with an overhanging horizontal branch 3.50 m above the release height of the arrow. Will the arrow go over or under the branch?

6. A rugby player passes the ball 7.00 m across the field, where it is caught at the same height as it left his hand. (a) At what angle was the ball thrown if its initial speed was 12.0 m/s, assuming that the smaller of the two possible angles was used? (b) What other angle gives the same range, and why would it not be used? (c) How long did this pass take?

7. Verify the ranges for the projectiles in Figure 5 (a) for θ = 45º and the given initial velocities.

8. Verify the ranges shown for the projectiles in Figure 5(b) for an initial velocity of 50 m/s at the given initial angles.

9. The cannon on a battleship can fire a shell a maximum distance of 32.0 km. (a) Calculate the initial velocity of the shell. (b) What maximum height does it reach? (At its highest, the shell is above 60% of the atmosphere—but air resistance is not really negligible as assumed to make this problem easier.) (c) The ocean is not flat, because the Earth is curved. Assume that the radius of the Earth is 6.37 × 10 3 . How many meters lower will its surface be 32.0 km from the ship along a horizontal line parallel to the surface at the ship? Does your answer imply that error introduced by the assumption of a flat Earth in projectile motion is significant here?

10. An arrow is shot from a height of 1.5 m toward a cliff of height H . It is shot with a velocity of 30 m/s at an angle of 60º above the horizontal. It lands on the top edge of the cliff 4.0 s later. (a) What is the height of the cliff? (b) What is the maximum height reached by the arrow along its trajectory? (c) What is the arrow’s impact speed just before hitting the cliff?

11. In the standing broad jump, one squats and then pushes off with the legs to see how far one can jump. Suppose the extension of the legs from the crouch position is 0.600 m and the acceleration achieved from this position is 1.25 times the acceleration due to gravity, g . How far can they jump? State your assumptions. (Increased range can be achieved by swinging the arms in the direction of the jump.)

12. The world long jump record is 8.95 m (Mike Powell, USA, 1991). Treated as a projectile, what is the maximum range obtainable by a person if he has a take-off speed of 9.5 m/s? State your assumptions.

13. Serving at a speed of 170 km/h, a tennis player hits the ball at a height of 2.5 m and an angle θ below the horizontal. The service line is 11.9 m from the net, which is 0.91 m high. What is the angle θ such that the ball just crosses the net? Will the ball land in the service box, whose out line is 6.40 m from the net?

14. A football quarterback is moving straight backward at a speed of 2.00 m/s when he throws a pass to a player 18.0 m straight downfield. (a) If the ball is thrown at an angle of 25º relative to the ground and is caught at the same height as it is released, what is its initial speed relative to the ground? (b) How long does it take to get to the receiver? (c) What is its maximum height above its point of release?

15. Gun sights are adjusted to aim high to compensate for the effect of gravity, effectively making the gun accurate only for a specific range. (a) If a gun is sighted to hit targets that are at the same height as the gun and 100.0 m away, how low will the bullet hit if aimed directly at a target 150.0 m away? The muzzle velocity of the bullet is 275 m/s. (b) Discuss qualitatively how a larger muzzle velocity would affect this problem and what would be the effect of air resistance.

16. An eagle is flying horizontally at a speed of 3.00 m/s when the fish in her talons wiggles loose and falls into the lake 5.00 m below. Calculate the velocity of the fish relative to the water when it hits the water.

17. An owl is carrying a mouse to the chicks in its nest. Its position at that time is 4.00 m west and 12.0 m above the center of the 30.0 cm diameter nest. The owl is flying east at 3.50 m/s at an angle 30.0º below the horizontal when it accidentally drops the mouse. Is the owl lucky enough to have the mouse hit the nest? To answer this question, calculate the horizontal position of the mouse when it has fallen 12.0 m.

18. Suppose a soccer player kicks the ball from a distance 30 m toward the goal. Find the initial speed of the ball if it just passes over the goal, 2.4 m above the ground, given the initial direction to be 40º above the horizontal.

19. Can a goalkeeper at her/ his goal kick a soccer ball into the opponent’s goal without the ball touching the ground? The distance will be about 95 m. A goalkeeper can give the ball a speed of 30 m/s.

20. The free throw line in basketball is 4.57 m (15 ft) from the basket, which is 3.05 m (10 ft) above the floor. A player standing on the free throw line throws the ball with an initial speed of 7.15 m/s, releasing it at a height of 2.44 m (8 ft) above the floor. At what angle above the horizontal must the ball be thrown to exactly hit the basket? Note that most players will use a large initial angle rather than a flat shot because it allows for a larger margin of error. Explicitly show how you follow the steps involved in solving projectile motion problems.

21. In 2007, Michael Carter (U.S.) set a world record in the shot put with a throw of 24.77 m. What was the initial speed of the shot if he released it at a height of 2.10 m and threw it at an angle of 38.0º above the horizontal? (Although the maximum distance for a projectile on level ground is achieved at 45º when air resistance is neglected, the actual angle to achieve maximum range is smaller; thus, 38º will give a longer range than 45º in the shot put.)

22. A basketball player is running at 5.00 m/s directly toward the basket when he jumps into the air to dunk the ball. He maintains his horizontal velocity. (a) What vertical velocity does he need to rise 0.750 m above the floor? (b) How far from the basket (measured in the horizontal direction) must he start his jump to reach his maximum height at the same time as he reaches the basket?

23. A football player punts the ball at a 45º angle. Without an effect from the wind, the ball would travel 60.0 m horizontally. (a) What is the initial speed of the ball? (b) When the ball is near its maximum height it experiences a brief gust of wind that reduces its horizontal velocity by 1.50 m/s. What distance does the ball travel horizontally?

24. Prove that the trajectory of a projectile is parabolic, having the form [latex]y=\text{ax}+{\text{bx}}^{2}\\[/latex]. To obtain this expression, solve the equation [latex]x={v}_{0x}t\\[/latex] for t and substitute it into the expression for [latex]y={v}_{0y}t-\left(1/2\right){\text{gt}}^{2}\\[/latex]. (These equations describe the x and y positions of a projectile that starts at the origin.) You should obtain an equation of the form [latex]y=\text{ax}+{\text{bx}}^{2}\\[/latex] where a and b are constants.

25. Derive [latex]R=\frac{{{v}_{0}}^{2}\text{\sin}{2\theta }_{0}}{g}\\[/latex] for the range of a projectile on level ground by finding the time t at which y becomes zero and substituting this value of t into the expression for x – x 0 , noting that R = x – x 0 .

26. Unreasonable Results (a) Find the maximum range of a super cannon that has a muzzle velocity of 4.0 km/s. (b) What is unreasonable about the range you found? (c) Is the premise unreasonable or is the available equation inapplicable? Explain your answer. (d) If such a muzzle velocity could be obtained, discuss the effects of air resistance, thinning air with altitude, and the curvature of the Earth on the range of the super cannon.

27. Construct Your Own Problem Consider a ball tossed over a fence. Construct a problem in which you calculate the ball’s needed initial velocity to just clear the fence. Among the things to determine are; the height of the fence, the distance to the fence from the point of release of the ball, and the height at which the ball is released. You should also consider whether it is possible to choose the initial speed for the ball and just calculate the angle at which it is thrown. Also examine the possibility of multiple solutions given the distances and heights you have chosen.

## Selected Solutions to Problems & Exercises

1. x = 1.30 m × 10 2 , y = 30.9 m

3. (a) 3.50 s (b) 28.6 m/s (c) 34.3 m/s (d) 44.7 m/s, 50.2º below horizontal

5. (a) 18.4º (b) The arrow will go over the branch.

7. [latex]R=\frac{{{{v}_{0}}}^{}}{\sin{2\theta }_{0}g}\\[/latex]

For θ = 45º, [latex]R=\frac{{{{v}_{0}}}^{2}}{g}\\[/latex]

R = 91.9 m for v 0 = 30 m/s; R = 163 m for v 0 ; R = 255 m for v 0 = 50 m/s

9. (a) 560 m/s (b) 800 × 10 3 m (c) 80.0 m. This error is not significant because it is only 1% of the answer in part (b).

11. 1.50 m, assuming launch angle of 45º

13. θ =6.1º. Yes, the ball lands at 5.3 m from the net

15. (a) −0.486 m (b) The larger the muzzle velocity, the smaller the deviation in the vertical direction, because the time of flight would be smaller. Air resistance would have the effect of decreasing the time of flight, therefore increasing the vertical deviation.

17. 4.23 m. No, the owl is not lucky; he misses the nest.

19. No, the maximum range (neglecting air resistance) is about 92 m.

21. 15.0 m/s

23. (a) 24.2 m/s (b) The ball travels a total of 57.4 m with the brief gust of wind.

25. [latex]y-{y}_{0}=0={v}_{0y}t-\frac{1}{2}{gt}^{2}=\left({v}_{0}\sin\theta\right)t-\frac{1}{2}{gt}^{2}\\[/latex] ,

so that [latex]t=\frac{2\left({v}_{0}\sin\theta \right)}{g}\\[/latex]

[latex]x-{x}_{0}={v}_{0x}t=\left({v}_{0}\cos\theta \right)t=R\\[/latex], and substituting for t gives:

[latex]R={v}_{0}\cos\theta \left(\frac{{2v}_{0}\sin\theta}{g}\right)=\frac{{{2v}_{0}}^{2}\sin\theta \cos\theta }{g}\\[/latex]

since [latex]2\sin\theta \cos\theta =\sin 2\theta\\[/latex], the range is:

[latex]R=\frac{{{v}_{0}}^{2}\sin 2\theta }{g}\\[/latex].

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Home A Level Kinematics (A Level) Steps To Solve Projectile Motion Questions

## Steps To Solve Projectile Motion Questions

- Speed, Velocity and Acceleration
- Signs For Motion, Displacement And Acceleration
- Reading Kinematics Graphs
- Equations of Motion
- Air Resistance
- Projectile Motion
- Make a clearly labelled sketch of the projectile motion.
- If you are given the initial velocity, resolve it into its x and y components.
- Analyze the horizontal and vertical motion separately.
- Decide on your sign convention. (Are you taking upwards or downwards as positive?)
- Recall: Acceleration is always taken as 9.81 m s -2 and is directed downwards. When an object is at its highest point of motion, its vertical velocity is always zero. The horizontal component of a projectile remains unchanged throughout the flight.
- Apply the relevant kinematics equations. (Think for a minute before jumping into the equations. A little planning goes a long way. Remember to take sign conventions into consideration!)

Back To Kinematics

Back To Kinematics (A Level)

Back To A Level Topic List

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## Mini Physics

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## 1 thought on “Steps To Solve Projectile Motion Questions”

i love the way you give your steps but it does not come with diagram you people should try and give diagram in your steps to enable fast learning

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PHYSICS CALCULATIONS

## Problems of Projectile Motion

Table of Contents

## Solutions to the Problems of Projectile Motion

Here are 11 solved problems of projectile motion to help you understand how to tackle any related question:

A stone is shot out from a catapult with an initial velocity of 65 meters per second at an elevation of 60 degrees. Find

a. Time of flight

b. Maximum Height attained

c. The range

Initial velocity, u = 65 m/s

The angle of elevation, θ = 60 0

Gravitational acceleration , g = 10 m/s 2

a. To find the time of the flight, we will apply the following methods:

Time of flight, T = ?

The formula for solving a problem on time of flight, T is (2usinθ) / g

We will now insert our data into the above formula

Since T = (2usinθ) / g

T = (2 x 65 x sin60 0 ) / 10 = (130 x 0.866) / 10 = 112.58 / 10 = 11.258 = 11.3 seconds

Therefore, the time of flight is 11.3 seconds

b. Here is how to calculate the maximum height

Maximum Height, H max = ?

The formula for calculating the maximum height is

Maximum height, H max = (u 2 sin 2 θ) / 2g

We can now substitute our formula with the data

The maximum height, H max = (65 2 sin 2 60 0 ) / 2 x 10

H max = (4225 x (sin60 0 ) 2 ) / 2 x 10

And the above equation will give us

H max = (4225 x 0.75) / 20

H max = 3168.75 / 20 = 158.4375 m

Our maximum height can be approximated into

H max = 158 m

Therefore, our maximum height is 158 meters .

c. We will find the range

The range, R = ?

Here is the formula that will help us find the range

The range, R = (u 2 sin2θ) / g

The range, R = (65 2 sin2(60 0 ) / g

And the above expression will give us

R = (4225sin120 0 ) / 10 = (4225 x 0.866) / 10 = 3658.85 / 10 = 365.885 m

Therefore, the range of the stone shot out of the catapult is 365.885 meters .

A bullet is fired horizontally with a velocity of 40 meters per second from the top of a building 50 meters high. How far from the feet of the building will the bullet be assumed to touch the ground? (Take the gravitational acceleration, g = 10 m/s 2 ).

Horizontal velocity = Initial velocity = u x = 40 m/s

Height of the building, H = 50 m

Time of flight, T = √ (2H / g)

T = √ (2x 50 / 10) = √ (100 / 10) = √ 10 = 3.162277 = 3 seconds

The range, R = uT = 40 x 3 = 120 m

Therefore, the bullet will cover 120 meters from the foot of the building to the ground.

A tennis ball is thrown vertically upwards from the front with a velocity of 50 m/s. Calculate

a. The maximum height reached

b. The time it takes to reach the maximum height

c. The time of the flight

Initial velocity, u = 50 m/s

a. Here is how to find the maximum height of the tennis ball

The formula for finding the maximum height is

H max = u 2 / 2g

After inserting our data into the formula H max = u 2 / 2g, we will have

Maximum height, H max = (50) 2 / 2 x 10 = 2,500 / 20 = 125 m

Therefore, the maximum height reached by the tennis ball is 125 meters

b. The time taken to reach the maximum height can be calculated this way:

The time it takes to reach the maximum height, t = ?

The formula to apply is t = u / g

We will apply our data to the formula t = u / g

Time, t = 50 / 10 = 5 seconds

c. Time of flight will be

T = 2u / g = (2 x 50) / 10 = 100 / 10 = 10 seconds.

A bullet is fired from a point making an angle of 36 degrees to the horizontal. The initial velocity of the bullet is 43 m/s. Find

a. The greatest height reached

b. The time taken to reach the maximum height

[Take gravitational acceleration, g = 9.8 m/s 2 ]

The angle made, θ = 36 0

Initial velocity, u = 43 m/s

Gravitational acceleration, g = 9.8 m/s 2

a. We can calculate the maximum height, H max this way

The maximum height, H max = ?

Apply the data into H max = (u 2 sin 2 θ) / 2g to obtain

H max = (43 2 sin 2 36 0 ) / 2 x 9.8

The maximum height, H max = (1849 x 0.34549) / 19.6 = 638.8 / 19.6 = 32.59 = 33 m

Therefore, the maximum height reached is 33 meters.

b. The time of flight will be

Time of flight, T = usinθ / g = 43sin36 0 / 9.8 = (43 x 0.59) / 9.8 = 25.37 / 9.8 = 2.589 s

Therefore, the time of flight is 2.6 seconds

A projectile is fired with an initial velocity of 100 m/s at an angle of 30 degrees with the horizontal. Calculate:

a. The time of flight

b. The maximum height attained

The initial velocity of the projectile, u = 100 m/s

Angle of inclination, θ = 30 0

Gravitational acceleration, g = 10 m/s 2

To find the time of flight, we will follow the method below:

we will apply

Time of flight, T = (2usinθ) / g

We will now put our data into the formula

T = (2 x 100 x sin30 0 ) / g = (200 x 0.5) / 10 = 100 /10 = 10 s

Therefore, the time of flight is10 seconds

To find the maximum height attained, we will use

We can now substitute our formula with our data

H max = (100 2 x sin 2 30 0 ) / (2 x 10)

H max = (10,000 x (0.5) 2 ) / 20 = (10,000 x 0.25) / 20 = 2,500 / 20 = 125 m

Therefore, the maximum height attained by the projectile is 125 meters

Now, we are going to calculate the range

After inserting our data into the above formula, we will get

R = (100 2 sin2 x 30 0 ) / 10 = (10,000 x sin60 0 ) / 10 = (10,000 x 0.866) / 10

And we will have

R = 8660 / 10 = 866 m

Therefore, our range is eight hundred and sixty-six (866) meters.

A bomber on a military mission is flying horizontally at a height of 3000 meters above the ground at 60 kilometers per minute. It drops a bomb on a target on the ground. Determine the acute angle between the vertical and the line joining the bomber and the target at the instant the bomb is released.

The horizontal velocity of the bomber aircraft = 60 km/min = (60 x 1000) / 60 = 60,000 / 60 = 1000 m/s

Initial velocity, u = 0

Height above the ground, h = 3000 m

The first formula is

h = ut + (1/2)gt 2 to find t

The second formula

Distance covered in time t, s = horizontal velocity x time to find s

The third formula is

tanθ = s / h to find the angle of inclination (θ)

We will need to insert our data into the first formula to find t

h = ut + (1/2)gt 2

Which can be rewritten as

3000 = (1/2) x 10 x t 2 [Remember that u = 0]

By making t subject of the formula, we will have

(3000 x 2) / 10 = t 2

We can now take the square root of both sides to obtain

And our final answer for the time will be

Time, t = 24.5 seconds

We will apply our second formula

Distance, s = horizontal velocity x time = 1000 x 24.5 = 24,500 meters

We will now proceed with our third formula

tanθ = s / h = 24,500 / 3000 = 8.2

Hence, we divide both sides by tan to obtain a tan inverse of 8.2

θ = tan -1 8.2 = 83.047 0

We can now approximate our answer into

Therefore, the angle of inclination is eighty-three (83) degrees.

A ball is projected horizontally from the top of a hill with a velocity of 30 meters per second. If it reaches the ground 5 seconds later. The height of the ball is

The initial velocity of the ball, u = 0

Time it takes to reach the ground, t = 5 s

Maximum height of the ball, h = ut + (1/2)gt 2

To solve the problem, insert your data into the above formula

h = ut + (1/2)gt 2 = 0 x 5 + (1/2) x 10 x 5 2 = (1/2) x 10 x 25 = (1/2) x 250 = 250 / 2 = 125 meter

Therefore, the height of the hill is 125 meters

A stone is projected at an angle of 60 degrees and an initial velocity of 20 meters per second. Determine the time of flight.

The angle of inclination of the stone, θ = 60 0

Initial velocity of the stone, u = 20 m/s

Gravitational acceleration, g = 10 ms -2

The formula for calculating the time of flight, T = (2usinθ) / g

We will put our data into the formula T = (2usinθ) / g

T = (2usinθ) / g = T = (2 x 20 x sin60 0 ) / 10 = (40 x 0.866) / 10 = 34.64 / 10 = 3.464 s

Therefore, the time of flight is 3.46 seconds.

A body is projected upward at an angle of 30 degrees with the horizontal at an initial speed of 200 meters per second. In how many seconds will it reach the ground? How far from the point of projection will it strike?

Angle of projection, θ = 200 m/s

Horizontal initial speed, u = 200 m/s

The question wants us to solve

Number of seconds for the body to reach ground = Time of flight, T = ?

We were also asked to find out how far from the point of projection will the body strike the ground. Hence, we are going to find the range.

First formula to find the time of flight, T = (2usinθ) / g

Second formula to find the range, R = (u 2 sin2θ) / g

To find the time of flight, we will insert our data into the first formula T = (2usinθ) / g

Time of flight, T = (2usinθ) / g = (2 x 200 x sin30 0 ) / 10

T = (400 x 0.5) / 10 = 200 / 10 = 20 s

Therefore, the time of flight is 20 seconds.

By applying the second formula R = (u 2 sin2θ) / g and inserting our data, we will have

R = (u 2 sin2θ) / g = (200 2 x sin (2×30 0 )) / 10

Which implies that

R = (40,000 x sin60 0 ) / 10 = (40,000 x 0.866) / 10

The range, R = 34,640 / 10

Therefore, the range covered is 3,464 meters

A tennis ball projected at an angle θ attains a range of R = 78 meters. If the velocity imparted to the ball by the racket is 30 meters per second, calculate the angle θ.

Range, R = 78 m

The velocity imparted to the ball by racket = initial velocity, u = 30 m/s

The angle θ = ?

We will use the formula R = (u 2 sin2θ) / g

Substitute your formula with the data

78 = (30 2 sin2θ) / 10

780 = (900sin2θ)

Sin2θ = 780 / 900

Sin2θ = 0.867

After dividing both sides by sin, will now have

2θ = sin -1 0.867

Which is equal to

2θ = 60.11 0

After approximating 60.11 0 to 60 0 . Therefore,

θ = 60 0 / 2 = 30 0

Therefore, the angle θ is 30 degrees

An aeroplane, flying in a straight line at a constant height of 500 meters with a speed of 200 meters per second drops a food package. The package takes a time t to reach the ground and travels a horizontal distance d in doing so. Taking g as 10 meters per second square, and ignoring air resistance, what are the values of t and d?

The height, h = 500 m

Speed of the aeroplane = 200 m/s

The time it takes the package to reach the ground, t = ?

Distance covered by the package to reach the ground, d = ?

First formula: We will apply t = √2h/g to find time (t)

Second Formula: We will use s = speed of the aeroplane x time it takes the package to drop on the ground [where s = distance covered by the package to reach the ground]

By applying our first formula t = √2h/g and inserting our data, we will obtain

t = √((2×500) / 10) = √1000/10 = √100 = 10 seconds

Therefore, the time it takes for the package to reach the ground is 10 seconds.

We will now apply the second formula to calculate the distance covered

Remember that s = speed of the aeroplane x time

s = 200 x 10 = 2000 m

Therefore, the distance covered by the package to reach the ground is 2000 meters

You may also like to read:

How to Calculate Maximum height

What are Prefixes in Physics

How to Find Displacement in Physics

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- 3.4 Projectile Motion
- Introduction to Science and the Realm of Physics, Physical Quantities, and Units
- 1.1 Physics: An Introduction
- 1.2 Physical Quantities and Units
- 1.3 Accuracy, Precision, and Significant Figures
- 1.4 Approximation
- Section Summary
- Conceptual Questions
- Problems & Exercises
- Introduction to One-Dimensional Kinematics
- 2.1 Displacement
- 2.2 Vectors, Scalars, and Coordinate Systems
- 2.3 Time, Velocity, and Speed
- 2.4 Acceleration
- 2.5 Motion Equations for Constant Acceleration in One Dimension
- 2.6 Problem-Solving Basics for One-Dimensional Kinematics
- 2.7 Falling Objects
- 2.8 Graphical Analysis of One-Dimensional Motion
- Introduction to Two-Dimensional Kinematics
- 3.1 Kinematics in Two Dimensions: An Introduction
- 3.2 Vector Addition and Subtraction: Graphical Methods
- 3.3 Vector Addition and Subtraction: Analytical Methods
- 3.5 Addition of Velocities
- Introduction to Dynamics: Newton’s Laws of Motion
- 4.1 Development of Force Concept
- 4.2 Newton’s First Law of Motion: Inertia
- 4.3 Newton’s Second Law of Motion: Concept of a System
- 4.4 Newton’s Third Law of Motion: Symmetry in Forces
- 4.5 Normal, Tension, and Other Examples of Forces
- 4.6 Problem-Solving Strategies
- 4.7 Further Applications of Newton’s Laws of Motion
- 4.8 Extended Topic: The Four Basic Forces—An Introduction
- Introduction: Further Applications of Newton’s Laws
- 5.1 Friction
- 5.2 Drag Forces
- 5.3 Elasticity: Stress and Strain
- Introduction to Uniform Circular Motion and Gravitation
- 6.1 Rotation Angle and Angular Velocity
- 6.2 Centripetal Acceleration
- 6.3 Centripetal Force
- 6.4 Fictitious Forces and Non-inertial Frames: The Coriolis Force
- 6.5 Newton’s Universal Law of Gravitation
- 6.6 Satellites and Kepler’s Laws: An Argument for Simplicity
- Introduction to Work, Energy, and Energy Resources
- 7.1 Work: The Scientific Definition
- 7.2 Kinetic Energy and the Work-Energy Theorem
- 7.3 Gravitational Potential Energy
- 7.4 Conservative Forces and Potential Energy
- 7.5 Nonconservative Forces
- 7.6 Conservation of Energy
- 7.8 Work, Energy, and Power in Humans
- 7.9 World Energy Use
- Introduction to Linear Momentum and Collisions
- 8.1 Linear Momentum and Force
- 8.2 Impulse
- 8.3 Conservation of Momentum
- 8.4 Elastic Collisions in One Dimension
- 8.5 Inelastic Collisions in One Dimension
- 8.6 Collisions of Point Masses in Two Dimensions
- 8.7 Introduction to Rocket Propulsion
- Introduction to Statics and Torque
- 9.1 The First Condition for Equilibrium
- 9.2 The Second Condition for Equilibrium
- 9.3 Stability
- 9.4 Applications of Statics, Including Problem-Solving Strategies
- 9.5 Simple Machines
- 9.6 Forces and Torques in Muscles and Joints
- Introduction to Rotational Motion and Angular Momentum
- 10.1 Angular Acceleration
- 10.2 Kinematics of Rotational Motion
- 10.3 Dynamics of Rotational Motion: Rotational Inertia
- 10.4 Rotational Kinetic Energy: Work and Energy Revisited
- 10.5 Angular Momentum and Its Conservation
- 10.6 Collisions of Extended Bodies in Two Dimensions
- 10.7 Gyroscopic Effects: Vector Aspects of Angular Momentum
- Introduction to Fluid Statics
- 11.1 What Is a Fluid?
- 11.2 Density
- 11.3 Pressure
- 11.4 Variation of Pressure with Depth in a Fluid
- 11.5 Pascal’s Principle
- 11.6 Gauge Pressure, Absolute Pressure, and Pressure Measurement
- 11.7 Archimedes’ Principle
- 11.8 Cohesion and Adhesion in Liquids: Surface Tension and Capillary Action
- 11.9 Pressures in the Body
- Introduction to Fluid Dynamics and Its Biological and Medical Applications
- 12.1 Flow Rate and Its Relation to Velocity
- 12.2 Bernoulli’s Equation
- 12.3 The Most General Applications of Bernoulli’s Equation
- 12.4 Viscosity and Laminar Flow; Poiseuille’s Law
- 12.5 The Onset of Turbulence
- 12.6 Motion of an Object in a Viscous Fluid
- 12.7 Molecular Transport Phenomena: Diffusion, Osmosis, and Related Processes
- Introduction to Temperature, Kinetic Theory, and the Gas Laws
- 13.1 Temperature
- 13.2 Thermal Expansion of Solids and Liquids
- 13.3 The Ideal Gas Law
- 13.4 Kinetic Theory: Atomic and Molecular Explanation of Pressure and Temperature
- 13.5 Phase Changes
- 13.6 Humidity, Evaporation, and Boiling
- Introduction to Heat and Heat Transfer Methods
- 14.2 Temperature Change and Heat Capacity
- 14.3 Phase Change and Latent Heat
- 14.4 Heat Transfer Methods
- 14.5 Conduction
- 14.6 Convection
- 14.7 Radiation
- Introduction to Thermodynamics
- 15.1 The First Law of Thermodynamics
- 15.2 The First Law of Thermodynamics and Some Simple Processes
- 15.3 Introduction to the Second Law of Thermodynamics: Heat Engines and Their Efficiency
- 15.4 Carnot’s Perfect Heat Engine: The Second Law of Thermodynamics Restated
- 15.5 Applications of Thermodynamics: Heat Pumps and Refrigerators
- 15.6 Entropy and the Second Law of Thermodynamics: Disorder and the Unavailability of Energy
- 15.7 Statistical Interpretation of Entropy and the Second Law of Thermodynamics: The Underlying Explanation
- Introduction to Oscillatory Motion and Waves
- 16.1 Hooke’s Law: Stress and Strain Revisited
- 16.2 Period and Frequency in Oscillations
- 16.3 Simple Harmonic Motion: A Special Periodic Motion
- 16.4 The Simple Pendulum
- 16.5 Energy and the Simple Harmonic Oscillator
- 16.6 Uniform Circular Motion and Simple Harmonic Motion
- 16.7 Damped Harmonic Motion
- 16.8 Forced Oscillations and Resonance
- 16.10 Superposition and Interference
- 16.11 Energy in Waves: Intensity
- Introduction to the Physics of Hearing
- 17.2 Speed of Sound, Frequency, and Wavelength
- 17.3 Sound Intensity and Sound Level
- 17.4 Doppler Effect and Sonic Booms
- 17.5 Sound Interference and Resonance: Standing Waves in Air Columns
- 17.6 Hearing
- 17.7 Ultrasound
- Introduction to Electric Charge and Electric Field
- 18.1 Static Electricity and Charge: Conservation of Charge
- 18.2 Conductors and Insulators
- 18.3 Coulomb’s Law
- 18.4 Electric Field: Concept of a Field Revisited
- 18.5 Electric Field Lines: Multiple Charges
- 18.6 Electric Forces in Biology
- 18.7 Conductors and Electric Fields in Static Equilibrium
- 18.8 Applications of Electrostatics
- Introduction to Electric Potential and Electric Energy
- 19.1 Electric Potential Energy: Potential Difference
- 19.2 Electric Potential in a Uniform Electric Field
- 19.3 Electrical Potential Due to a Point Charge
- 19.4 Equipotential Lines
- 19.5 Capacitors and Dielectrics
- 19.6 Capacitors in Series and Parallel
- 19.7 Energy Stored in Capacitors
- Introduction to Electric Current, Resistance, and Ohm's Law
- 20.1 Current
- 20.2 Ohm’s Law: Resistance and Simple Circuits
- 20.3 Resistance and Resistivity
- 20.4 Electric Power and Energy
- 20.5 Alternating Current versus Direct Current
- 20.6 Electric Hazards and the Human Body
- 20.7 Nerve Conduction–Electrocardiograms
- Introduction to Circuits and DC Instruments
- 21.1 Resistors in Series and Parallel
- 21.2 Electromotive Force: Terminal Voltage
- 21.3 Kirchhoff’s Rules
- 21.4 DC Voltmeters and Ammeters
- 21.5 Null Measurements
- 21.6 DC Circuits Containing Resistors and Capacitors
- Introduction to Magnetism
- 22.1 Magnets
- 22.2 Ferromagnets and Electromagnets
- 22.3 Magnetic Fields and Magnetic Field Lines
- 22.4 Magnetic Field Strength: Force on a Moving Charge in a Magnetic Field
- 22.5 Force on a Moving Charge in a Magnetic Field: Examples and Applications
- 22.6 The Hall Effect
- 22.7 Magnetic Force on a Current-Carrying Conductor
- 22.8 Torque on a Current Loop: Motors and Meters
- 22.9 Magnetic Fields Produced by Currents: Ampere’s Law
- 22.10 Magnetic Force between Two Parallel Conductors
- 22.11 More Applications of Magnetism
- Introduction to Electromagnetic Induction, AC Circuits and Electrical Technologies
- 23.1 Induced Emf and Magnetic Flux
- 23.2 Faraday’s Law of Induction: Lenz’s Law
- 23.3 Motional Emf
- 23.4 Eddy Currents and Magnetic Damping
- 23.5 Electric Generators
- 23.6 Back Emf
- 23.7 Transformers
- 23.8 Electrical Safety: Systems and Devices
- 23.9 Inductance
- 23.10 RL Circuits
- 23.11 Reactance, Inductive and Capacitive
- 23.12 RLC Series AC Circuits
- Introduction to Electromagnetic Waves
- 24.1 Maxwell’s Equations: Electromagnetic Waves Predicted and Observed
- 24.2 Production of Electromagnetic Waves
- 24.3 The Electromagnetic Spectrum
- 24.4 Energy in Electromagnetic Waves
- Introduction to Geometric Optics
- 25.1 The Ray Aspect of Light
- 25.2 The Law of Reflection
- 25.3 The Law of Refraction
- 25.4 Total Internal Reflection
- 25.5 Dispersion: The Rainbow and Prisms
- 25.6 Image Formation by Lenses
- 25.7 Image Formation by Mirrors
- Introduction to Vision and Optical Instruments
- 26.1 Physics of the Eye
- 26.2 Vision Correction
- 26.3 Color and Color Vision
- 26.4 Microscopes
- 26.5 Telescopes
- 26.6 Aberrations
- Introduction to Wave Optics
- 27.1 The Wave Aspect of Light: Interference
- 27.2 Huygens's Principle: Diffraction
- 27.3 Young’s Double Slit Experiment
- 27.4 Multiple Slit Diffraction
- 27.5 Single Slit Diffraction
- 27.6 Limits of Resolution: The Rayleigh Criterion
- 27.7 Thin Film Interference
- 27.8 Polarization
- 27.9 *Extended Topic* Microscopy Enhanced by the Wave Characteristics of Light
- Introduction to Special Relativity
- 28.1 Einstein’s Postulates
- 28.2 Simultaneity And Time Dilation
- 28.3 Length Contraction
- 28.4 Relativistic Addition of Velocities
- 28.5 Relativistic Momentum
- 28.6 Relativistic Energy
- Introduction to Quantum Physics
- 29.1 Quantization of Energy
- 29.2 The Photoelectric Effect
- 29.3 Photon Energies and the Electromagnetic Spectrum
- 29.4 Photon Momentum
- 29.5 The Particle-Wave Duality
- 29.6 The Wave Nature of Matter
- 29.7 Probability: The Heisenberg Uncertainty Principle
- 29.8 The Particle-Wave Duality Reviewed
- Introduction to Atomic Physics
- 30.1 Discovery of the Atom
- 30.2 Discovery of the Parts of the Atom: Electrons and Nuclei
- 30.3 Bohr’s Theory of the Hydrogen Atom
- 30.4 X Rays: Atomic Origins and Applications
- 30.5 Applications of Atomic Excitations and De-Excitations
- 30.6 The Wave Nature of Matter Causes Quantization
- 30.7 Patterns in Spectra Reveal More Quantization
- 30.8 Quantum Numbers and Rules
- 30.9 The Pauli Exclusion Principle
- Introduction to Radioactivity and Nuclear Physics
- 31.1 Nuclear Radioactivity
- 31.2 Radiation Detection and Detectors
- 31.3 Substructure of the Nucleus
- 31.4 Nuclear Decay and Conservation Laws
- 31.5 Half-Life and Activity
- 31.6 Binding Energy
- 31.7 Tunneling
- Introduction to Applications of Nuclear Physics
- 32.1 Diagnostics and Medical Imaging
- 32.2 Biological Effects of Ionizing Radiation
- 32.3 Therapeutic Uses of Ionizing Radiation
- 32.4 Food Irradiation
- 32.5 Fusion
- 32.6 Fission
- 32.7 Nuclear Weapons
- Introduction to Particle Physics
- 33.1 The Yukawa Particle and the Heisenberg Uncertainty Principle Revisited
- 33.2 The Four Basic Forces
- 33.3 Accelerators Create Matter from Energy
- 33.4 Particles, Patterns, and Conservation Laws
- 33.5 Quarks: Is That All There Is?
- 33.6 GUTs: The Unification of Forces
- Introduction to Frontiers of Physics
- 34.1 Cosmology and Particle Physics
- 34.2 General Relativity and Quantum Gravity
- 34.3 Superstrings
- 34.4 Dark Matter and Closure
- 34.5 Complexity and Chaos
- 34.6 High-temperature Superconductors
- 34.7 Some Questions We Know to Ask
- A | Atomic Masses
- B | Selected Radioactive Isotopes
- C | Useful Information
- D | Glossary of Key Symbols and Notation

## Learning Objectives

By the end of this section, you will be able to:

- Identify and explain the properties of a projectile, such as acceleration due to gravity, range, maximum height, and trajectory.
- Determine the location and velocity of a projectile at different points in its trajectory.
- Apply the principle of independence of motion to solve projectile motion problems.

Projectile motion is the motion of an object thrown or projected into the air, subject to only the acceleration of gravity. The object is called a projectile , and its path is called its trajectory . The motion of falling objects, as covered in Problem-Solving Basics for One-Dimensional Kinematics , is a simple one-dimensional type of projectile motion in which there is no horizontal movement. In this section, we consider two-dimensional projectile motion, such as that of a football or other object for which air resistance is negligible .

The most important fact to remember here is that motions along perpendicular axes are independent and thus can be analyzed separately. This fact was discussed in Kinematics in Two Dimensions: An Introduction , where vertical and horizontal motions were seen to be independent. The key to analyzing two-dimensional projectile motion is to break it into two motions, one along the horizontal axis and the other along the vertical. (This choice of axes is the most sensible, because acceleration due to gravity is vertical—thus, there will be no acceleration along the horizontal axis when air resistance is negligible.) As is customary, we call the horizontal axis the x -axis and the vertical axis the y -axis. Figure 3.34 illustrates the notation for displacement, where s s is defined to be the total displacement and x x and y y are its components along the horizontal and vertical axes, respectively. The magnitudes of these vectors are s , x , and y . (Note that in the last section we used the notation A A to represent a vector with components A x A x and A y A y . If we continued this format, we would call displacement s s with components s x s x and s y s y . However, to simplify the notation, we will simply represent the component vectors as x x and y y .)

Of course, to describe motion we must deal with velocity and acceleration, as well as with displacement. We must find their components along the x - and y -axes, too. We will assume all forces except gravity (such as air resistance and friction, for example) are negligible. The components of acceleration are then very simple: a y = – g = – 9.80 m /s 2 a y = – g = – 9.80 m /s 2 . (Note that this definition assumes that the upwards direction is defined as the positive direction. If you arrange the coordinate system instead such that the downwards direction is positive, then acceleration due to gravity takes a positive value.) Because gravity is vertical, a x = 0 a x = 0 . Both accelerations are constant, so the kinematic equations can be used.

## Review of Kinematic Equations (constant a a )

Given these assumptions, the following steps are then used to analyze projectile motion:

Step 1. Resolve or break the motion into horizontal and vertical components along the x- and y-axes. These axes are perpendicular, so A x = A cos θ A x = A cos θ and A y = A sin θ A y = A sin θ are used. The magnitude of the components of displacement s s along these axes are x x and y. y. The magnitudes of the components of the velocity v v are v x = v cos θ v x = v cos θ and v y = v sin θ, v y = v sin θ, where v v is the magnitude of the velocity and θ θ is its direction, as shown in Figure 3.35 . Initial values are denoted with a subscript 0, as usual.

Step 2. Treat the motion as two independent one-dimensional motions, one horizontal and the other vertical. The kinematic equations for horizontal and vertical motion take the following forms:

Step 3. Solve for the unknowns in the two separate motions—one horizontal and one vertical. Note that the only common variable between the motions is time t t . The problem solving procedures here are the same as for one-dimensional kinematics and are illustrated in the solved examples below.

Step 4. Recombine the two motions to find the total displacement s s and velocity v v . Because the x - and y -motions are perpendicular, we determine these vectors by using the techniques outlined in the Vector Addition and Subtraction: Analytical Methods and employing A = A x 2 + A y 2 A = A x 2 + A y 2 and θ = tan − 1 ( A y / A x ) θ = tan − 1 ( A y / A x ) in the following form, where θ θ is the direction of the displacement s s and θ v θ v is the direction of the velocity v v :

Total displacement and velocity

## Example 3.4

A fireworks projectile explodes high and away.

During a fireworks display, a shell is shot into the air with an initial speed of 70.0 m/s at an angle of 75.0º 75.0º above the horizontal, as illustrated in Figure 3.36 . The fuse is timed to ignite the shell just as it reaches its highest point above the ground. (a) Calculate the height at which the shell explodes. (b) How much time passed between the launch of the shell and the explosion? (c) What is the horizontal displacement of the shell when it explodes?

Because air resistance is negligible for the unexploded shell, the analysis method outlined above can be used. The motion can be broken into horizontal and vertical motions in which a x = 0 a x = 0 and a y = – g a y = – g . We can then define x 0 x 0 and y 0 y 0 to be zero and solve for the desired quantities.

## Solution for (a)

By “height” we mean the altitude or vertical position y y above the starting point. The highest point in any trajectory, called the apex, is reached when v y = 0 v y = 0 . Since we know the initial and final velocities as well as the initial position, we use the following equation to find y y :

Because y 0 y 0 and v y v y are both zero, the equation simplifies to

Solving for y y gives

Now we must find v 0 y v 0 y , the component of the initial velocity in the y -direction. It is given by v 0 y = v 0 sin θ v 0 y = v 0 sin θ , where v 0 y v 0 y is the initial velocity of 70.0 m/s, and θ 0 = 75.0º θ 0 = 75.0º is the initial angle. Thus,

## Discussion for (a)

Note that because up is positive, the initial velocity is positive, as is the maximum height, but the acceleration due to gravity is negative. Note also that the maximum height depends only on the vertical component of the initial velocity, so that any projectile with a 67.6 m/s initial vertical component of velocity will reach a maximum height of 233 m (neglecting air resistance). The numbers in this example are reasonable for large fireworks displays, the shells of which do reach such heights before exploding. In practice, air resistance is not completely negligible, and so the initial velocity would have to be somewhat larger than that given to reach the same height.

## Solution for (b)

As in many physics problems, there is more than one way to solve for the time to the highest point. In this case, the easiest method is to use y = y 0 + 1 2 ( v 0 y + v y ) t y = y 0 + 1 2 ( v 0 y + v y ) t . Because y 0 y 0 is zero, this equation reduces to simply

Note that the final vertical velocity, v y v y , at the highest point is zero. Thus,

## Discussion for (b)

This time is also reasonable for large fireworks. When you are able to see the launch of fireworks, you will notice several seconds pass before the shell explodes. (Another way of finding the time is by using y = y 0 + v 0 y t − 1 2 gt 2 y = y 0 + v 0 y t − 1 2 gt 2 , and solving the quadratic equation for t t .)

## Solution for (c)

Because air resistance is negligible, a x = 0 a x = 0 and the horizontal velocity is constant, as discussed above. The horizontal displacement is horizontal velocity multiplied by time as given by x = x 0 + v x t x = x 0 + v x t , where x 0 x 0 is equal to zero:

where v x v x is the x -component of the velocity, which is given by v x = v 0 cos θ 0 . v x = v 0 cos θ 0 . Now,

The time t t for both motions is the same, and so x x is

## Discussion for (c)

The horizontal motion is a constant velocity in the absence of air resistance. The horizontal displacement found here could be useful in keeping the fireworks fragments from falling on spectators. Once the shell explodes, air resistance has a major effect, and many fragments will land directly below.

In solving part (a) of the preceding example, the expression we found for y y is valid for any projectile motion where air resistance is negligible. Call the maximum height y = h y = h ; then,

This equation defines the maximum height of a projectile and depends only on the vertical component of the initial velocity.

## Defining a Coordinate System

It is important to set up a coordinate system when analyzing projectile motion. One part of defining the coordinate system is to define an origin for the x x and y y positions. Often, it is convenient to choose the initial position of the object as the origin such that x 0 = 0 x 0 = 0 and y 0 = 0 y 0 = 0 . It is also important to define the positive and negative directions in the x x and y y directions. Typically, we define the positive vertical direction as upwards, and the positive horizontal direction is usually the direction of the object’s motion. When this is the case, the vertical acceleration, g g , takes a negative value (since it is directed downwards towards the Earth). However, it is occasionally useful to define the coordinates differently. For example, if you are analyzing the motion of a ball thrown downwards from the top of a cliff, it may make sense to define the positive direction downwards since the motion of the ball is solely in the downwards direction. If this is the case, g g takes a positive value.

## Example 3.5

Calculating projectile motion: hot rock projectile.

Kilauea in Hawaii is the world’s most continuously active volcano. Very active volcanoes characteristically eject red-hot rocks and lava rather than smoke and ash. Suppose a large rock is ejected from the volcano with a speed of 25.0 m/s and at an angle 35.0º 35.0º above the horizontal, as shown in Figure 3.37 . The rock strikes the side of the volcano at an altitude 20.0 m lower than its starting point. (a) Calculate the time it takes the rock to follow this path. (b) What are the magnitude and direction of the rock’s velocity at impact?

Again, resolving this two-dimensional motion into two independent one-dimensional motions will allow us to solve for the desired quantities. The time a projectile is in the air is governed by its vertical motion alone. We will solve for t t first. While the rock is rising and falling vertically, the horizontal motion continues at a constant velocity. This example asks for the final velocity. Thus, the vertical and horizontal results will be recombined to obtain v v and θ v θ v at the final time t t determined in the first part of the example.

While the rock is in the air, it rises and then falls to a final position 20.0 m lower than its starting altitude. We can find the time for this by using

If we take the initial position y 0 y 0 to be zero, then the final position is y = − 20 .0 m . y = − 20 .0 m . Now the initial vertical velocity is the vertical component of the initial velocity, found from v 0 y = v 0 sin θ 0 v 0 y = v 0 sin θ 0 = ( 25 . 0 m/s 25 . 0 m/s )( sin 35.0º sin 35.0º ) = 14 . 3 m/s 14 . 3 m/s . Substituting known values yields

Rearranging terms gives a quadratic equation in t t :

This expression is a quadratic equation of the form at 2 + bt + c = 0 at 2 + bt + c = 0 , where the constants are a = 4.90 a = 4.90 , b = – 14.3 b = – 14.3 , and c = – 20.0. c = – 20.0. Its solutions are given by the quadratic formula:

This equation yields two solutions: t = 3.96 t = 3.96 and t = – 1.03 t = – 1.03 . (It is left as an exercise for the reader to verify these solutions.) The time is t = 3.96 s t = 3.96 s or – 1.03 s – 1.03 s . The negative value of time implies an event before the start of motion, and so we discard it. Thus,

The time for projectile motion is completely determined by the vertical motion. So any projectile that has an initial vertical velocity of 14.3 m/s and lands 20.0 m below its starting altitude will spend 3.96 s in the air.

From the information now in hand, we can find the final horizontal and vertical velocities v x v x and v y v y and combine them to find the total velocity v v and the angle θ 0 θ 0 it makes with the horizontal. Of course, v x v x is constant so we can solve for it at any horizontal location. In this case, we chose the starting point since we know both the initial velocity and initial angle. Therefore:

The final vertical velocity is given by the following equation:

where v 0y v 0y was found in part (a) to be 14 . 3 m/s 14 . 3 m/s . Thus,

To find the magnitude of the final velocity v v we combine its perpendicular components, using the following equation:

which gives

The direction θ v θ v is found from the equation:

The negative angle means that the velocity is 50 . 1º 50 . 1º below the horizontal. This result is consistent with the fact that the final vertical velocity is negative and hence downward—as you would expect because the final altitude is 20.0 m lower than the initial altitude. (See Figure 3.37 .)

One of the most important things illustrated by projectile motion is that vertical and horizontal motions are independent of each other. Galileo was the first person to fully comprehend this characteristic. He used it to predict the range of a projectile. On level ground, we define range to be the horizontal distance R R traveled by a projectile. Galileo and many others were interested in the range of projectiles primarily for military purposes—such as aiming cannons. However, investigating the range of projectiles can shed light on other interesting phenomena, such as the orbits of satellites around the Earth. Let us consider projectile range further.

How does the initial velocity of a projectile affect its range? Obviously, the greater the initial speed v 0 v 0 , the greater the range, as shown in Figure 3.38 (a). The initial angle θ 0 θ 0 also has a dramatic effect on the range, as illustrated in Figure 3.38 (b). For a fixed initial speed, such as might be produced by a cannon, the maximum range is obtained with θ 0 = 45º θ 0 = 45º . This is true only for conditions neglecting air resistance. If air resistance is considered, the maximum angle is approximately 38º 38º . Interestingly, for every initial angle except 45º 45º , there are two angles that give the same range—the sum of those angles is 90º 90º . The range also depends on the value of the acceleration of gravity g g . The lunar astronaut Alan Shepherd was able to drive a golf ball a great distance on the Moon because gravity is weaker there. The range R R of a projectile on level ground for which air resistance is negligible is given by

where v 0 v 0 is the initial speed and θ 0 θ 0 is the initial angle relative to the horizontal. The proof of this equation is left as an end-of-chapter problem (hints are given), but it does fit the major features of projectile range as described.

When we speak of the range of a projectile on level ground, we assume that R R is very small compared with the circumference of the Earth. If, however, the range is large, the Earth curves away below the projectile and acceleration of gravity changes direction along the path. The range is larger than predicted by the range equation given above because the projectile has farther to fall than it would on level ground. (See Figure 3.39 .) If the initial speed is great enough, the projectile goes into orbit. This possibility was recognized centuries before it could be accomplished. When an object is in orbit, the Earth curves away from underneath the object at the same rate as it falls. The object thus falls continuously but never hits the surface. These and other aspects of orbital motion, such as the rotation of the Earth, will be covered analytically and in greater depth later in this text.

Once again we see that thinking about one topic, such as the range of a projectile, can lead us to others, such as the Earth orbits. In Addition of Velocities , we will examine the addition of velocities, which is another important aspect of two-dimensional kinematics and will also yield insights beyond the immediate topic.

## PhET Explorations

Projectile motion.

Blast a Buick out of a cannon! Learn about projectile motion by firing various objects. Set the angle, initial speed, and mass. Add air resistance. Make a game out of this simulation by trying to hit a target.

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Access for free at https://openstax.org/books/college-physics-2e/pages/1-introduction-to-science-and-the-realm-of-physics-physical-quantities-and-units

- Authors: Paul Peter Urone, Roger Hinrichs
- Publisher/website: OpenStax
- Book title: College Physics 2e
- Publication date: Jul 13, 2022
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