3.3

Projectile Motion

The biggest thrill in baseball is a home run. The motion of the ball on its curving path into the stands is a common type of two-dimensional motion called “projectile motion.” A good description of such motion can often be obtained with the assumption that air resistance is absent.

CONCEPTS AT A GLANCE Following the approach outlined in Figure 3-6, we consider the horizontal and vertical parts of the motion separately. In the horizontal or x direction, the moving object (the projectile) does not slow down in the absence of air resistance. Thus, the x component of the velocity remains constant at its initial value or , and the x component of the acceleration is . In the vertical or y direction, however, the projectile experiences the effect of gravity. As a result, the y component of the velocity vy is not constant, but changes. The y component of the acceleration ay is the downward acceleration due to gravity. If the path or trajectory of the projectile is near the earth’s surface, ay has a magnitude of 9.80 m/s2. In this text, then, the phrase “projectile motion” means that and ay equals the acceleration due to gravity, as the Concepts-at-a-Glance chart in Figure 3-8 summarizes. Example 3 and other examples in this section illustrate how the equations of kinematics are applied to projectile motion.

Figure 3-8 CONCEPTS AT A GLANCE In projectile motion, the horizontal or x component of the acceleration is zero, and the vertical or y component of the acceleration is the acceleration due to gravity. The water shooting upward in this Parisian fountain exhibits projectile motion, to the extent that the effects of air resistance are negligible. The arcing paths that the water follows in the photograph are characteristic of projectile motion. (Benjamin Rondel/Corbis Images)

| Example 3 | | | A Falling Care Package

Figure 3-9 shows an airplane moving horizontally with a constant velocity of +115 m/s at an altitude of 1050 m. The directions to the right and upward have been chosen as the positive directions. The plane releases a “care package” that falls to the ground along a curved trajectory. Ignoring air resistance, determine the time required for the package to hit the ground.

Figure 3-9 The package falling from the plane is an example of projectile motion, as Example 3 and Example 4 discuss.

Reasoning The time required for the package to hit the ground is the time it takes for the package to fall through a vertical distance of 1050 m. In falling, it moves to the right, as well as downward, but these two parts of the motion occur independently. Therefore, we can focus solely on the vertical part. We note that the package is moving initially in the horizontal or x direction, not in the y direction, so that . Furthermore, when the package hits the ground, the y component of its displacement is , as the drawing shows. The acceleration is that due to gravity, so . These data are summarized as follows:

y-Direction Data
y | ay | vy | v0y | t
−1050 m | −9.80 m/s2 |  | 0 m/s | ?

Problem solving insight The variables y, ay, vy, and v0y are scalar components. Therefore, an algebraic sign (+ or −) must be included with each one to denote direction.

With these data, Equation 3.5b () can be used to find the fall time.
Solution Since , it follows from Equation 3.5b that and

The freely falling package in Example 3 picks up vertical speed on the way downward. The horizontal component of the velocity, however, retains its initial value of throughout the entire descent. Since the plane also travels at a constant horizontal velocity of +115 m/s, it remains directly above the falling package. The pilot always sees the package directly beneath the plane, as the dashed vertical lines in Figure 3-9 show. This result is a direct consequence of the fact that the package has no acceleration in the horizontal direction. In reality, air resistance would slow down the package, and it would not remain directly beneath the plane during the descent. Figure 3-10 further clarifies this point by illustrating what happens to two packages that are released simultaneously from the same height. Package A is dropped from a stationary balloon and falls straight downward toward the ground, since it has no horizontal velocity component (). Package B, on the other hand, is given an initial velocity component of in the horizontal direction, as in Example 3, and follows the path shown in the figure. Both packages hit the ground at the same time.

Figure 3-10 Package A and package B are released simultaneously at the same height and strike the ground at the same time because their y variables (y, ay, and v0y) are the same.

Not only do the packages in Figure 3-10 reach the ground at the same time, but the y components of their velocities are also equal at all points on the way down. However, package B does hit the ground with a greater speed than does package A. Remember, speed is the magnitude of the velocity vector, and the velocity of B has an x component, whereas the velocity of A does not. The magnitude and direction of the velocity vector for package B at the instant just before the package hits the ground is computed in Example 4.

A N A L Y Z I N G M U L T I P L E - C O N C E P T P R O B L E M S

| Example 4 | | | The Velocity of the Care Package

Figure 3-9 shows a care package falling from a plane, and Figure 3-10 shows this package as package B. As in Example 3, the directions to the right and upward are chosen as the positive directions, and the plane is moving horizontally with a constant velocity of +115 m/s at an altitude of 1050 m. Ignoring air resistance, find the magnitude v and the directional angle of the final velocity vector that the package has just before it strikes the ground.
Reasoning Figures 3-9 and 3-10 show the final velocity vector, which has components vx and vy and a magnitude v. The vector is directed at an angle below the horizontal or x direction. We note the right triangle formed by the vector and its components. The hypotenuse of the triangle is the magnitude of the velocity, and the components are the other two sides. As in Example 3, we can use the Pythagorean theorem to express the magnitude or speed v in terms of the components vx and vy, and we can use trigonometry to determine the directional angle .
Knowns and Unknowns The data for this problem are listed in the table that follows:

Description | Symbol | Value | Comment
Explicit Data, x Direction
x component of initial velocity | v0x | +115 m/s | Package has plane’s horizontal velocity at instant of release
Implicit Data, x Direction
x component of acceleration | ax | 0 m/s2 | No horizontal acceleration, since air resistance ignored
Explicit Data, y Direction
y component of displacement | y | −1050 m | Negative, since upward is positive and package falls downward
Implicit Data, y Direction
y component of initial velocity | v0y | 0 m/s | Package traveling horizontally in x direction at instant of release, not in y direction
y component of acceleration | ay | −9.80 m/s2 | Acceleration vector for gravity points downward in the negative direction
Unknown Variables
Magnitude of final velocity | v | ? |
Direction of final velocity |  | ? |

Modeling the Problem

| Step 1 Final Velocity
| Using the Pythagorean theorem to express the speed v in terms of the components vx and vy (see Figure 3-9 or 3-10), we obtain Equation 1a at the right. Furthermore, in a right triangle, the cosine of an angle is the side adjacent to the angle divided by the hypotenuse. With this in mind, we see in Figure 3-9 or Figure 3-10 that the directional angle is given by Equation 1b at the right.

| (1a)

| (1b)

| Step 2 The Components of the Final Velocity
| Reference to the table of knowns and unknowns shows that, in the x direction, values are available for the kinematic variables v0x and ax. Since the acceleration ax is zero, the final velocity component vx remains unchanged from its initial value of v0x, so we have

In the y direction, values are available for y, v0y, and ay, so that we can determine the final velocity component vy using Equation 3.6b from the equations of kinematics:

| (3.6b)

These results for vx and vy can be substituted into Equations 1a and 1b, as shown at the right.

Solution Algebraically combining the results of each step, we find that

Problem solving insight The speed of a projectile at any location along its path is the magnitude v of its velocity at that location: . Both the horizontal and vertical velocity components contribute to the speed.

With the data given for the kinematic variables in the x and y directions, we find that the magnitude and direction of the final velocity of the package are

Related Homework: Problems 27, 29 and 38

An important feature of projectile motion is that there is no acceleration in the horizontal, or x, direction. Conceptual Example 5 discusses an interesting implication of this feature.

| Conceptual Example 5 | | | I Shot a Bullet into the Air …

Suppose you are driving in a convertible with the top down. The car is moving to the right at a constant velocity. As Figure 3-11 illustrates, you point a rifle straight upward and fire it. In the absence of air resistance, where would the bullet land—behind you, ahead of you, or in the barrel of the rifle?

Figure 3-11 The car is moving with a constant velocity to the right, and the rifle is pointed straight up. In the absence of air resistance, a bullet fired from the rifle has no acceleration in the horizontal direction. As a result, the bullet would land back in the barrel of the rifle.

Reasoning and Solution If air resistance were present, it would slow down the bullet and cause it to land behind you, toward the rear of the car. However, air resistance is absent, so we must consider the bullet’s motion more carefully. Before the rifle is fired, the bullet, rifle, and car are moving together, so the bullet and rifle have the same horizontal velocity as the car. When the rifle is fired, the bullet is given an additional velocity component in the vertical direction; the bullet retains the velocity of the car as its initial horizontal velocity component, since the rifle is pointed straight up. Because there is no air resistance to slow it down, the bullet experiences no horizontal acceleration. Thus, the bullet’s horizontal velocity component does not change. It retains its initial value, and remains matched to that of the rifle and the car. As a result, the bullet remains directly above the rifle at all times and would fall directly back into the barrel of the rifle, as the drawing indicates. This situation is analogous to that in Figure 3-9, where the care package, as it falls, remains directly below the plane.
Related Homework: Problem 68

Often projectiles, such as footballs and baseballs, are sent into the air at an angle with respect to the ground. From a knowledge of the projectile’s initial velocity, a wealth of information can be obtained about the motion. For instance, Example 6 demonstrates how to calculate the maximum height reached by the projectile.

| Example 6 | | | The Height of a Kickoff

A placekicker kicks a football at an angle of above the horizontal axis, as Figure 3-12 shows. The initial speed of the ball is . Ignore air resistance and find the maximum height H that the ball attains.

Figure 3-12 A football is kicked with an initial speed of v0 at an angle of above the ground. The ball attains a maximum height H and a range R.

Reasoning The maximum height is a characteristic of the vertical part of the motion, which can be treated separately from the horizontal part. In preparation for making use of this fact, we calculate the vertical component of the initial velocity:

The vertical component of the velocity, vy, decreases as the ball moves upward. Eventually, at the maximum height H. The data below can be used in Equation 3.6b () to find the maximum height:

y-Direction Data
y | ay | vy | v0y | t
H = ? | −9.80 m/s2 | 0 m/s | +14 m/s |

Solution From Equation 3.6b, we find that

The height H depends only on the y variables; the same height would have been reached had the ball been thrown straight up with an initial velocity of .

Problem solving insight When a projectile reaches maximum height, the vertical component of its velocity is momentarily zero (). However, the horizontal component of its velocity is not zero.

It is also possible to find the total time or “hang time” during which the football in Figure 3-12 is in the air. Example 7 shows how to determine this time.

| Example 7 | | | The Time of Flight of a Kickoff

For the motion illustrated in Figure 3-12, ignore air resistance and use the data from Example 6 to determine the time of flight between kickoff and landing.

The physics of the “hang time” of a football.

Reasoning Given the initial velocity, it is the acceleration due to gravity that determines how long the ball stays in the air. Thus, to find the time of flight we deal with the vertical part of the motion. Since the ball starts at and returns to ground level, the displacement in the y direction is zero. The initial velocity component in the y direction is the same as that in Example 6; that is, . Therefore, we have

y-Direction Data
y | ay | vy | v0y | t
0 m | −9.80 m/s2 |  | +14 m/s | ?

The time of flight can be determined from Equation 3.5b ().
Solution Using Equation 3.5b, we find

There are two solutions to this equation. One is given by

The other is given by . The solution we seek is , because corresponds to the initial kickoff.

Another important feature of projectile motion is called the “range.” The range, as Figure 3-12 shows, is the horizontal distance traveled between launching and landing, assuming the projectile returns to the same vertical level at which it was fired. Example 8 shows how to obtain the range.

| Example 8 | | | The Range of a Kickoff

For the motion shown in Figure 3-12 and discussed in Example 6 and Example 7, ignore air resistance and calculate the range R of the projectile. Reasoning The range is a characteristic of the horizontal part of the motion. Thus, our starting point is to determine the horizontal component of the initial velocity:

Recall from Example 7 that the time of flight is . Since there is no acceleration in the x direction, vx remains constant, and the range is simply the product of and the time.
Solution The range is

The range in the previous example depends on the angle at which the projectile is fired above the horizontal. When air resistance is absent, the maximum range results when .

Need more practice?

Interactive LearningWare3.2
In 1971 astronaut Alan Shepard walked on the moon’s surface. In a moment of whimsy, he hit a golf ball, which was launched upward and followed the familiar trajectory. However, the trajectory differed from what it would have been on earth, because the acceleration due to gravity on the moon is about six times smaller than that on earth. Consider the same ball on earth, launched at the same angle, with the same speed. Find the ratio of (a) the maximum height of the moon-ball to the maximum height of the earth-ball and (b) the range of the moon-ball to the range of the earth-ball.
Related Homework: Problems 19 and 23

Check Your Understanding 3

A projectile is fired into the air, and it follows the parabolic path shown in the drawing, landing on the right. There is no air resistance. At any instant, the projectile has a velocity and an acceleration . Which one or more of the drawings could not represent the directions for and at any point on the trajectory? (The answer is given at the end of the book.)

Background: The fundamental nature of projectile motion lies at the heart of this question, as does the nature of gravity.
For similar questions (including calculational counterparts), consult Self-Assessment Test 3.1. The test is described at the end of this section.

The examples considered thus far have used information about the initial location and velocity of a projectile to determine the final location and velocity. Example 9 deals with the opposite situation and illustrates how the final parameters can be used with the equations of kinematics to determine the initial parameters.

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| Example 9 | | | A Home Run

A baseball player hits a home run, and the ball lands in the left-field seats, 7.5 m above the point at which it was hit. It lands with a velocity of 36 m/s at an angle of 28° below the horizontal (see Figure 3-13.) The positive directions are upward and to the right in the drawing. Ignoring air resistance, find the magnitude and direction of the initial velocity with which the ball leaves the bat.

Figure 3-13 The velocity and location of the baseball upon landing can be used to determine its initial velocity, as Example 9 illustrates.

Reasoning Just after the ball is hit, its initial velocity has a magnitude v0 and components of v0x and v0y and is directed at an angle above the horizontal or x direction. Figure 3-13 shows the initial velocity vector and its components. As usual, we will use the Pythagorean theorem to relate v0 to v0x and v0y and will use trigonometry to determine .
Knowns and Unknowns The data for this problem are listed in the table that follows:

Description | Symbol | Value | Comment
Explicit Data
y component of displacement | y | +7.5 m | Positive, since upward is positive and ball lands above its starting point
Magnitude of final velocity | v | 36 m/s |
Direction of final velocity |  | 28° | Below the horizontal (see Figure 3-13)
Implicit Data
x component of acceleration | ax | 0 m/s | No horizontal acceleration, since air resistance ignored
y component of acceleration | ay | −9.80 m/s2 | Acceleration vector for gravity points downward in the negative direction
Unknown Variables
Magnitude of initial velocity | v0 | ? |
Direction of initial velocity |  | ? |

Modeling the Problem

| Step 1 Initial Velocity
| The magnitude v0 of the initial velocity can be related to its components v0x and v0y by using the Pythagorean theorem, since the components are perpendicular to one another. This leads to Equation 1a at the right. Referring to Figure 3-13, we can also use trigonometry to express the directional angle in terms of the components v0x and v0y. Thus, we obtain Equation 1b at the right.

| (1a)

| (1b)

| Step 2 x Component of the Initial Velocity
| To obtain v0x, we note that the acceleration ax is zero, since air resistance is being ignored. With no acceleration in the x direction, v0x remains unchanged throughout the motion of the ball. Thus, v0x must equal vx, the x component of the ball’s final velocity. We have, then, that

This result can be substituted into Equations 1a and 1b, as shown at the right.

| Step 3 y Component of the Initial Velocity
| In contrast to the argument in Step 2 The Components of the Final Velocity, v0y does not equal vy, the y component of the ball’s final velocity, since the ball accelerates in the vertical direction. However, we can use Equation 3.6b from the equations of kinematics to determine v0y:

| (3.6b)

The plus sign is chosen for the square root, since the ball’s initial velocity component v0y points upward in Figure 3-13. In Equation 3.6b, vy can be written as

where the minus sign is present because vy points downward in the −y direction in Figure 3-13. We find, then, that

This result can also be substituted into Equations 1a and 1b, as shown at the right.

Solution Algebraically combining the results of each step, we find that

With the data given in the table of knowns and unknowns, we find that the magnitude and direction of the ball’s initial velocity are

Related Homework: Problem 34 and 66
In projectile motion, the magnitude of the acceleration due to gravity affects the trajectory in a significant way. For example, a baseball or a golf ball would travel much farther and higher on the moon than on the earth, when launched with the same initial velocity. The reason is that the moon’s gravity is only about one-sixth as strong as the earth’s.
Section 2.6 points out that certain types of symmetry with respect to time and speed are present for freely falling bodies. These symmetries are also found in projectile motion, since projectiles are falling freely in the vertical direction. In particular, the time required for a projectile to reach its maximum height H is equal to the time spent returning to the ground. In addition, Figure 3-14 shows that the speed v of the object at any height above the ground on the upward part of the trajectory is equal to the speed v at the same height on the downward part. Although the two speeds are the same, the velocities are different, because they point in different directions. Conceptual Example 10 shows how to use this type of symmetry in your reasoning.

Figure 3-14 The speed v of a projectile at a given height above the ground is the same on the upward and downward parts of the trajectory. The velocities are different, however, since they point in different directions.

| Conceptual Example 10 | | | Two Ways to Throw a Stone

From the top of a cliff overlooking a lake, a person throws two stones. The stones have identical initial speeds v0, but stone 1 is thrown downward at an angle below the horizontal, while stone 2 is thrown upward at the same angle above the horizontal, as Figure 3-15 shows. Neglect air resistance and decide which stone, if either, strikes the water with the greater velocity.

Figure 3-15 Two stones are thrown off the cliff with identical initial speeds v0, but at equal angles that are below and above the horizontal. Conceptual Example 10 compares the velocities with which the stones hit the water below.

Reasoning and Solution We might guess that stone 1, being hurled downward, would strike the water with the greater velocity. To show that this is not true, let’s follow the path of stone 2 as it rises to its maximum height and falls back to earth. Notice point P in the drawing, where stone 2 returns to its initial height; here the speed of stone 2 is v0, but its velocity is directed at an angle below the horizontal. This is exactly the type of projectile symmetry illustrated in Figure 3-14. At this point, then, stone 2 has a velocity that is identical to the velocity with which stone 1 is thrown downward from the top of the cliff. From this point on, the velocity of stone 2 changes in exactly the same way as that for stone 1, so both stones strike the water with the same velocity.
Related Homework: Problems 21 and 71

In all the examples in this section, the projectiles follow a curved trajectory. In general, if the only acceleration is that due to gravity, the shape of the path can be shown to be a parabola.

Test your understanding of the material in Sections 3.1–3.3:

· | Motion in Two Dimensions
· | The Equations of Kinematics for Constant Acceleration
· | Projectile Motion

Copyright © 2007 John Wiley & Sons, Inc. All rights reserved.