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Key Termprojectile motion

Two-Dimensional MotionPreviously, we showed how quantities such as displacement and velocity were vectors that could be resolved into components. In this section, these components will be used to understand and predict the motion of objects thrown into the air.

Use of components avoids vector multiplication.How can you know the displacement, velocity, and acceleration of a ball at any point in time during its flight? All of the kinematic equations could be rewritten in terms of vector quantities. However, when an object is propelled into the air in a direction other than straight up or down, the velocity, acceleration, and displacement of the object do not all point in the same direction. This makes the vector forms of the equations difficult to solve.

One way to deal with these situations is to avoid using the complicated vector forms of the equations altogether. Instead, apply the technique of resolving vectors into components. Then you can apply the simpler one-dimensional forms of the equations for each component. Finally, you can recombine the components to determine the resultant.

Components simplify projectile motion.When a long jumper approaches his jump, he runs along a straight line, which can be called the x-axis. When he jumps, as shown in Figure 3.1, his velocity has both horizontal and vertical components. Movement in this plane can be depicted by using both the x- and y-axes.

Note that in Figure 3.2(b), a jumper’s velocity vector is resolved into its two vector components. This way, the jumper’s motion can be analyzed using the kinematic equations applied to one direction at a time.

Projectile Motion Objectives

Recognize examples of projectile motion.

Describe the path of a projectile as a parabola.

Resolve vectors into their components and apply the kinematic equations to solve problems involving projectile motion.

FIGURE 3.2

Motion of a Long Jumper When the long jumper is in the air, his velocity has both a horizontal and a vertical component.

FIGURE 3.1

Components of a Long Jumper’s Velocity (a) A long jumper’s velocity while sprinting along the runway can be represented by a horizontal vector. (b) Once the jumper is airborne, the jumper’s velocity at any instant can be described by the components of the velocity.

Two-Dimensional Motion and Vectors 93

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Differentiated Instruction

Preview VocabularyVisual Vocabulary Use two tennis balls to illustrate the difference between a projectile and a falling object. Drop one of the balls and point out how it moves down to the floor along a straight line. Then throw the other ball toward the wall and point out how this ball moves on a curved path until it hits the target. An object moving in a straight line toward the ground is falling. An object moving on curved path from its point of origin until it reaches a target point is called a projectile.

FIGURE 3.1 Tell students that the long jumper builds up speed in the x direc-tion and jumps, so there is also a component of speed in the y direction.

Ask Does the angle of takeoff matter to the jumper? Consider the difference between a very small angle (near 0°) and a larger angle (near 45°).

Answer: The angle matters because it affects how long the jumper stays off the ground and how far he goes horizontally while in the air.

Teaching TipOn the chalkboard, show examples of vector components and the kinematic equations. Show the simplification of the x-direction equations when the x component of acceleration is zero.

� Plan and Prepare

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TEACH FROM VISUALS

InclUsIonVisually impaired students may benefit from experiencing motions whose vectors have different components. Go outdoors or to an indoor location where students may move about safely, such as a large, empty lobby or gym. Have students jump straight up in place. Ask, how did you move? up, then down, or vertically This demonstrates motion in one dimension.

Next, have students jump forward a short distance. Again ask, how did you move? up and down, and also forward (horizontally) How many components were involved when you jumped up? one–the vertical direction How many components were involved when you jumped forward? two–vertical and horizontal In which case were you a projectile? when I jumped forward


sEcTIon 3

Path with air resistance

Path without air resistance

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projectile motion the curved path that an object follows when thrown, launched, or otherwise projected near the surface of Earth

Air Resistance Affects Projectile Motion (a) Without air resistance, the soccer ball would travel along a parabola. (b) With air resistance, the soccer ball would travel along a shorter path.

FIGURE 3.3

Did YOU Know?

The greatest distance a regulation-size baseball has ever been thrown is 135.9 m, by Glen Gorbous in 1957.

In this section, we will focus on the form of two-dimensional motion called projectile motion. Objects that are thrown or launched into the air and are subject to gravity are called projectiles. Some examples of projec-tiles are softballs, footballs, and arrows when they are projected through the air. Even a long jumper can be considered a projectile.

Projectiles follow parabolic trajectories.The path of a projectile is a curve called a parabola, as shown in Figure 3.3(a). Many people mistakenly think that projectiles eventually fall straight down in much the same way that a cartoon character does after running off a cliff. But if an object has an initial horizontal velocity, there will be horizontal motion throughout the flight of the projectile. Note that for the purposes of samples and exercises in this book, the horizontal velocity of projectiles will be considered constant. This velocity would not be constant if we accounted for air resistance. With air resistance, projec-tiles slow down as they collide with air particles, as shown in Figure 3.3(b).

Projectile motion is free fall with an initial horizontal velocity.To understand the motion a projectile undergoes, first examine Figure 3.4 on the following page. The red ball was dropped at the same instant the yellow ball was launched horizontally. If air resistance is disregarded, both balls hit the ground at the same time. By examining each ball’s position in relation to the horizontal lines and to one another, we see that the two balls fall at the same rate. This may seem impossible because one is given an initial velocity and the other begins from rest. But if the motion is analyzed one component at a time, it makes sense.

First, consider the red ball that falls straight down. It has no motion in the horizontal direction. In the vertical direction, it starts from rest (vy,i = 0 m/s) and proceeds in free fall. Thus, the kinematic equations from the chapter “Motion in One Dimension” can be applied to analyze the vertical motion of the falling ball, as shown on the next page. Note that on Earth’s surface the accel eration (ay) will equal −g (−9.81 m/s2) because the only vertical component of acceleration is free-fall acceleration. Note also that ∆y is negative.

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Differentiated Instruction

AIR REsIsTAncE Purpose Show the effects of air resistance on the flight of a projectile.

Materials rubber stopper, table-tennis ball

Procedure Toss the stopper at an angle of 45°. Have students note the trajec-tory. Sketch the parabolic path on the chalkboard. Throw the table-tennis ball at an angle of 45°. Have students note the trajectory. Sketch its path on the chalkboard. Have students compare the two trajectories. Students should note that the path of the table-tennis ball is not symmetrical but has a steeper descent, demonstrating the effect of air resistance.

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Demonstration

InclUsIonReview the general equation for a parabola, y = ax2 + bx + c, and illustrate a parabola on the board for clarity. Visual learners will benefit from time-lapse photography of a projectile in motion. Or consider using the classroom sink and a strobe light to illustrate the movement of water drops. You could begin by starting a slow drip from the faucet, darkening the room, and setting the strobe light up so that it shines through the drops.

Using your fingers or a stick, manipulate the water so that it arcs downward. This is only part of the parabola but students may be able to get the general idea of what occurs when a projectile passes through part of a parabola.

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Vertical Motion of a Projectile That Falls from Restvy,f = ay ∆t

vy,f2 = 2ay ∆y

∆y = 1 _ 2

ay(∆t)2

Now consider the components of motion of the yellow ball that is launched in Figure 3.4. This ball undergoes the same horizontal displace-ment during each time interval. This means that the ball’s horizontal velocity remains constant (if air resistance is assumed to be negligible). Thus, when the kinematic equations are used to analyze the horizontal motion of a projectile, the initial horizontal velocity is equal to the horizontal velocity throughout the projectile’s flight. A projectile’s hori-zontal motion is described by the following equation.

Horizontal Motion of a Projectilevx = vx,i = constant

∆x = vx ∆t

Next consider the initial motion of the launched yellow ball in Figure 3.4. Despite having an initial horizontal velocity, the launched ball has no initial velocity in the vertical direction. Just like the red ball that falls straight down, the launched yellow ball is in free fall. The vertical motion of the launched yellow ball is described by the same free-fall equations. In any time interval, the launched ball undergoes the same vertical displacement as the ball that falls straight down. For this reason, both balls reach the ground at the same time.

To find the velocity of a projectile at any point during its flight, find the vector that has the known components. Specifically, use the Pythago rean theorem to find the magnitude of the velocity, and use the tangent function to find the direction of the velocity.

FIGURE 3.4

PROJECTILE MOTION

Roll a ball off a table. At the instant the rolling ball leaves the table, drop a second ball from the same height above the floor. Do the two balls hit the floor at the same time?

Try varying the speed at which you roll the first ball off the table. Does varying the speed affect

whether the two balls strike the ground at the same time? Next roll one of the balls down a slope. Drop the other ball from the base of the slope at the instant the first ball leaves the slope. Which of the balls hits the ground first in this situation?

MATERIALS2 identical balls•

slope or ramp•

SAFETY Perform this experiment away from walls and furniture that can be damaged.

Vertical Motion of a Projectile This is a strobe photograph of two table-tennis balls released at the same time. Even though the yellow ball is given an initial horizontal velocity and the red ball is simply dropped, both balls fall at the same rate.


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Teaching TipExplain that the vertical motion of a projectile has negative acceleration. This is due to gravity. So, the vertical components of all motions of projectiles are affected by gravity and have negative acceleration.

TEAchER’s noTEsDropping the second ball as the first leaves the table is tricky. Students should try this several times in order to get the timing right. Holding the second ball just past the edge of the table and near the path of the first ball works well. You should point out the limitations of this QuickLab because of human reaction time.

QuickLab

PRE-APExplain that the horizontal motion of a projectile is independent of its vertical motion. Have students manipulate the equations given in the text to show that both equations are dependent on ∆t. In other words, show that the the vertical component has no effect on the horizontal motion and that the horizontal component has no effect on the vertical motion.


Projectiles Launched Horizontally

Sample Problem D The Royal Gorge Bridge in Colorado rises 321 m above the Arkansas River. Suppose you kick a rock horizontally off the bridge. The magnitude of the rock’s horizontal displacement is 45.0 m. Find the speed at which the rock was kicked.

ANALYZE Given: ∆y = –321 m ∆x = 45.0 m ay = –g = –9.81 m/s2

Unknown: vi = vx = ?

Diagram: The initial velocity vector of the rock has only a horizontal component. Choose the coordinate system oriented so that the positive y direction points upward and the positive x direction points to the right.

PLAN Choose an equation or situation:Because air resistance can be neglected, the rock’s horizontal velocity remains constant.

∆x = vx ∆t

Because there is no initial vertical velocity, the following equation applies.

∆y = 1 _ 2

ay(∆t)2

Rearrange the equations to isolate the unknowns:Note that the time interval is the same for the vertical and horizontal displacements, so the second equation can be rearranged to solvefor ∆t.

∆t = √ ��

2∆y _ a y

Next rearrange the first equation for vx, and substitute the above value of ∆t into the new equation.

vx = ∆x _ ∆t

= ( √ ��

a y

_ 2∆y

) ∆x

SOLVE Substitute the values into the equation and solve:

vx = √ �� –9.81 m/s2 __

(2) (–321 m) (45.0 m) = 5.56 m/s

CHECK YOUR ANSWER

To check your work, estimate the value of the time interval for ∆x and solve for ∆y. If vx is about 5.5 m/s and ∆x = 45 m, ∆t ≈ 8 s. If you use an approximate value of 10 m/s2 for g, ∆y ≈ –320 m, almost identical to the given value.

Continued

displacements, so the second equation can be rearranged to solvefor ∆

Next rearrange the first equation for the above value of

45.0 m

–321 m

PREMIUM CONTENT

Interactive DemoHMDScience.com

Tips and TricksThe value for vx can be either positive or negative because of the square root. Because the object is moving in what has been selected as the positive direction, you choose the positive answer.

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Problem Solving

DECONSTRUCTING PROBLEMSStudents may need to see the rearrangement involved in isolating the time interval. Step them through it as follows:

∆y = 1 _ 2 ay (∆t)2 Given

∆y _

1 _ 2 ay

= 1 _ 2 ay (∆t )2

_ 1 _ 2 ay

Divide both sides by 1 _ 2 ay.

2∆y _ a

y = (∆t)2 Simplify.

√ ��

2∆y _ a

y = ∆t Take square root

of each side.

Classroom PracticeProjectiles Launched Horizontally People in movies often jump from buildings into pools. If a person jumps horizontally from the 10th floor (30.0 m) to a pool that is 5.0 m away from the building, with what initial speed must the person jump?

Answer: 2.0 m/s

PROBLEM GUIDEDUse this guide to assign problems.SE = Student Edition TextbookPW = Sample Problem Set I (online)PB = Sample Problem Set II (online)Solving for:

vxSE Sample, 1–3;

Ch. Rvw. 31–32, 51aPW 6, 7*, 8*PB 9

∆x SE 4; Ch. Rvw. 33PW Sample, 1–2PB 8, 10

∆y SE Ch. Rvw. 51bPW 3–4, 5*PB Sample, 1–7

*Challenging Problem

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Projectiles Launched Horizontally (continued)

1. A baseball rolls off a 0.70 m high desk and strikes the floor 0.25 m away from the base of the desk. How fast was the ball rolling?

2. A cat chases a mouse across a 1.0 m high table. The mouse steps out of the way, and the cat slides off the table and strikes the floor 2.2 m from the edge of the table. When the cat slid off the table, what was its speed?

3. A pelican flying along a horizontal path drops a fish from a height of 4 m. The fish travels 8.0 m horizontally before it hits the water below. What is the pelican’s speed?

4. If the pelican in item 3 was traveling at the same speed but was only 2.7 m above the water, how far would the fish travel horizontally before hitting the water below?

Use components to analyze objects launched at an angle.Let us examine a case in which a projectile is launched at an angle to the horizontal, as shown in Figure 3.5. The projectile has an initial vertical component of velocity as well as a horizontal component of velocity.

Suppose the initial velocity vector makes an angle θ with the horizontal. Again, to analyze the motion of such a projectile, you must resolve the initial velocity vector into its components. The sine and cosine functions can be used to find the horizontal and vertical components of the initial velocity.

vx,i = vi cos θ and vy,i = vi sin θ

We can substitute these values for vx,i and vy,i into the kinematic equations to obtain a set of equations that can be used to analyze the motion of a projectile launched at an angle.

Projectiles Launched at an Anglevx = vx,i = vi cos θ = constant

∆x = (vi cos θ)∆t

vy,f = vi sin θ + ay ∆t

vy,f2 = vi

2 (sin θ)2 + 2ay ∆y

∆y = (vi sin θ)∆t + 1 _ 2

ay(∆t)2

As we have seen, the velocity of a projectile launched at an angle to the ground has both horizontal and vertical components. The vertical motion is similar to that of an object that is thrown straight up with an initial velocity.

Components of Initial Velocity An object is projected with an initial velocity, vi, at an angle of θ. Resolve the initial velocity into its x and y components. Then, the kinematic equations can be applied to describe the motion of the projectile throughout its flight.

FIGURE 3.5


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AnswersPractice D 1. 0.66 m/s

2. 4.9 m/s

3. 7.6 m/s

4. 5.6 m

FIGURE 3.5 Point out to students that from the time immediately after firing until it hits the ground, the projectile follows a parabolic path. Emphasize that the vector vi does not represent any part of the path of the projectile but only the direction and magnitude of its initial velocity.

Ask What is the acceleration of a projectile just before it hits the ground?

Answer: −9.81 m/s2, the same as at any other time during the flight

TEACH FROM VISUALS

REAlITy chEckStudents often memorize equations without taking the time to understand them. Encourage students to visualize how each of these equations is a form of one of the kinematic equations they learned earlier. Explain that solving projectile problems is done by applying the kinematic equations separately in each direction.


Projectiles Launched at an Angle

Sample Problem E A zookeeper finds an escaped monkey on a pole. While aiming her tranquilizer gun at the monkey, she kneels 10.0 m from the pole, which is 5.00 m high. The tip of her gun is 1.00 m above the ground. At the moment the zookeeper shoots, the monkey drops a banana. The dart travels at 50.0 m/s. Will the dart hit the monkey, the banana, or neither one?

ANALYZE Select a coordinate system.The positive y-axis points up, and the positive x-axis points along the ground toward the pole. Because the dart leaves the gun at a height of 1.00 m, the vertical distance is 4.00 m.

PLAN Use the inverse tangent function to find the angle of the dart with the x-axis.

θ = tan–1 ( ∆y

_ ∆x

) = tan–1 ( 4.00 m _ 10.0 m

) = 21.8°

Choose a kinematic equation to solve for time. Rearrange the equation for motion along the x-axis to isolate ∆t, theunknown, the time the dart takes to travel the horizontal distance.

∆x = (vi cos θ)∆t

∆t = ∆x _ vi cos θ

= 10.0 m __ (50.0 m/s)(cos 21.8°)

= 0.215 s

SOLVE Find out how far each object will fall during this time. Use the free-fall kinematic equation. For the banana, vi = 0. Thus:

∆yb = 1 _ 2

ay (∆t)2 = 1 _ 2

(–9.81 m/s2)(0.215 s)2 = –0.227 m

The dart has an initial vertical component of velocity of vi sin θ, so:

∆yd = (vi sin θ)∆t + 1 _ 2

ay(∆t)2

∆yd = (50.0 m/s)(sin 21.8°)(0.215 s) + 1 _ 2

(–9.81 m/s2)(0.215 s)2

∆yd = 3.99 m – 0.227 m = 3.76 m

Find the final height of both the banana and the dart.

yb, f = yb,i + ∆yb = 5.00 m + (−0.227 m) =

4.77 m above the ground

yd, f = yd,i + ∆yd = 1.00 m + 3.76 m =

4.76 m above the ground

The dart hits the banana. The slight difference is due to rounding.

Continued

TSI GraphicsHRW • Holt Physics

PH99PE-C03-003-013-A

10.0 m

4.00 m

1.00 m

Chapter 398



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Classroom PracticeProjectiles Launched at an Angle A golfer practices driving balls off a cliff and into the water below. The edge of the cliff is 15 m above the water. If the golf ball is launched at 51 m/s at an angle of 15° above the horizontal, how far does the ball travel horizontally before hitting the water? (See Appendix A for hints on solving quadratic equations.)

Answer: 1.7 × 102 m

PROBLEM guidE EUse this guide to assign problems.SE = Student Edition TextbookPW = Sample Problem Set I (online)PB = Sample Problem Set II (online)Solving for:

∆y/∆x SE Sample, 1–3; Ch. Rvw. 34a, 35–36, 55a, 56b, 59–60, 62a, 62c

PW 4b, 5, 7–8PB 6

νiSE 4; Ch. Rvw.

47a*, 48a, 49*, 56a, 61

PW Sample, 1–3, 4aPB 8, 10

∆t SE Ch. Rvw. 34b, 47*b, 55b, 62b

PB 7, 9

νfSE Ch. Rvw. 47c*,

48b

θ PW 6PB Sample, 1–5

*Challenging Problem

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PHYSICSSpec. Number PH 99 PE C03-003-010-ABoston Graphics, Inc.617.523.1333

200.0 m

= 30.0 m/splane

Reviewing Main Ideas

1. Which of the following exhibit parabolic motion?a. a flat rock skipping across the surface of a lakeb. a three-point shot in basketballc. a space shuttle while orbiting Earthd. a ball bouncing across a roome. a life preserver dropped from a stationary helicopter

2. During a thunderstorm, a tornado lifts a car to a height of 125 m above the ground. Increasing in strength, the tornado flings the car horizontally with a speed of 90.0 m/s. How long does the car take to reach the ground? How far horizontally does the car travel before hitting the ground?

Interpreting Graphics

3. An Alaskan rescue plane drops a package of emergency rations to a stranded party of explorers, as illustrated in Figure 3.6. The plane is traveling horizontally at 30.0 m/s at a height of 200.0 m above the ground.a. What horizontal distance does the package

fall before landing?b. Find the velocity of the package just before

it hits the ground.

1. In a scene in an action movie, a stuntman jumps from the top of one building to the top of another building 4.0 m away. After a running start, he leaps at a velocity of 5.0 m/s at an angle of 15° with respect to the flat roof. Will he make it to the other roof, which is 2.5 m lower than the building he jumps from?

2. A golfer hits a golf ball at an angle of 25.0° to the ground. If the golf ball covers a horizontal distance of 301.5 m, what is the ball’s maximum height? (Hint: At the top of its flight, the ball’s vertical velocity component will be zero.)

3. A baseball is thrown at an angle of 25° relative to the ground at a speed of 23.0 m/s. If the ball was caught 42.0 m from the thrower, how long was it in the air? How high above the thrower did the ball travel?

4. Salmon often jump waterfalls to reach their breeding grounds. One salmon starts 2.00 m from a waterfall that is 0.55 m tall and jumps at an angle of 32.0°. What must be the salmon’s minimum speed to reach the waterfall?

Dropping a Package

FIGURE 3.6

Projectiles Launched at an Angle (continued)


SECTION 3 FORMATIVE ASSESSMENT

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Answers to section Assessment

AnswersPractice E 1. yes, ∆y = −2.3 m

2. 35.1 m

3. 2.0 s; 4.8 m

4. 6.2 m/s

Assess Use the Formative Assessment on this page to evaluate student mastery of the section.

Reteach For students who need additional instruction, download the Section Study Guide.

Response to Intervention To reassess students’ mastery, use the Section Quiz, available to print or to take directly online at hMDscience.com.

� Assess and Reteach

1. a, b, d

2. 5.05 s; 454 m

3. a. 192 m

b. 69.4 m/s at 64.4° below the horizontal


section 3 projectile motion objectives · 8/2/2017 · baseball has ever been thrown is 135.9 m,...

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