2d kinematics projectiles relative...

2D KinematicsProjectiles

Relative Motion

Lana Sheridan

De Anza College

Jan 16, 2020

Last Time

• projectiles and principle equations

• height, time of flight, and range of a projectile

Overview

• trajectory equation

• projectile example

• introduce relative motion (?)

Projectiles

Last lecture we derived:

max height, h =(vi sin θi )

2

2g

And for cases where the projectile lands at the same height it waslaunched from:

time of flight, tfl =2vi sin θi

g

range, R =v2i sin(2θi )

g

Height and initial speed conceptual questionThree projectiles (A, B, and C) are launched with different initialspeeds so that they reach the same maximum height, as shown.List the projectiles in order of increasing time of flight.

100 CHAPTER 4 TWO-DIMENSIONAL KINEMATICS

positive y axis points vertically upward. What was the projec-tile’s launch angle with respect to the x axis if, at its highestpoint, its direction of motion has rotated (a) clockwise through50° or (b) counterclockwise through 30°? Explain.

Conceptual Exercises(Answers to odd-numbered Conceptual Exercises can be found in the back of the book.)

1. As you walk briskly down the street, you toss a small ball intothe air. If you want the ball to land in your hand when it comesback down, should you toss the ball (a) straight upward rela-tive to yourself, (b) in a forward direction, or (c) in a backwarddirection? Explain.

2. A certain projectile is launched with an initial speed At itshighest point its speed is Was the launch angle ofthe projectile (a) 30°, (b) 45°, or (c) 60°?

3. Two divers run horizontally off the edge of a low cliff. Diver 2runs with twice the speed of diver 1. When the divers hitthe water, is the horizontal distance covered by diver 2 (a) thesame as, (b) twice as much as, or (c) four times the horizontaldistance covered by diver 1?

4. Two youngsters dive off an overhang into a lake. Diver 1 dropsstraight down, diver 2 runs off the cliff with an initial horizon-tal speed Is the splashdown speed of diver 2 (a) greater than,(b) less than, or (c) equal to the splashdown speed of diver 1?

5. Three projectiles (A, B, and C) are launched with the same initialspeed but with different launch angles, as shown in Figure 4–12.List the projectiles in order of increasing (a) horizontal compo-nent of initial velocity and (b) time of flight. Indicate a tie with anequal sign.

v0.

12 v0.

v0.

6. Three projectiles (A, B, and C) are launched with different initialspeeds so that they reach the same maximum height, as shownin Figure 4–13. List the projectiles in order of increasing (a) ini-tial speed and (b) time of flight. Indicate a tie with an equal sign.

7. The penguin to the left in the accompanying photo is about toland an ice floe. Just before it lands, is its speed (a) greater than,(b) less than, or (c) the same as when it left the water? Explain.

11. Child 1 throws a snowball horizontally from the top of a roof;child 2 throws a snowball straight down. Once in flight, is theacceleration of snowball 2 (a) greater than, (b) equal to, or (c)less than the acceleration of snowball 1?

O x

C

B

A

y

▲ FIGURE 4–12 Conceptual Exercise 5

O x

A B C

y


This penguin behaves much like a projectile from the timeit leaves the water until it touches down on the ice.

(Conceptual Exercise 7)

Spee

d (m

/s)

15

10

5

0

1

2

3

TT

Time

12


the truck. Will the tomato hit your car or land on the road,assuming you continue moving with the same speed and direc-tion? Explain.

11. A projectile is launched from the origin of a coordinate systemwhere the positive x axis points horizontally to the right and the

8. A person flips a coin into the air and it lands on the ground afew feet away. If the person were to perform an identical coinflip on an elevator rising with constant speed, would the coin’stime of flight be (a) greater than, (b) less than, or (c) the same aswhen the person was at rest?

9. Suppose the elevator in the previous Exercise is rising with con-stant acceleration, rather than constant velocity. In this case,would the coin’s time of flight be (a) greater than, (b) less than,or (c) the same as when the person was at rest?

10. You throw a ball into the air with an initial speed of 10 m/s atan angle of 60° above the horizontal. The ball returns to thelevel from which it was thrown in the time T. Referring toFigure 4-14, which of the plots (1, 2, or 3) best represents thespeed of the ball as a function of time? What is wrong with theother two plots?

WALKMC04_0131536311.QXD 3/7/07 5:41 PM Page 100

(A) A, B, C

(B) C, B, A

(C) B, C, A

(D) all the same1Walker, “Physics”, page 106, prob 28.

Projectile Trajectory

Suppose we want to know the height of a projectile (relative to itslaunch point) in terms of its x coordinate. Suppose it is launchedat an angle θ above the horizontal, with initial velocity vi .

For the x-direction:

x = vi cos θt ⇒ t =x

vi cos θ

y -direction:

y = vi sin θt −1

2gt2

Substituting for t gives:

y = (tan θ)x −g

2v2i cos2 θ

x2

Projectile Motion Example: #25, page 103

Problems 103

of mass, which we will define in Chapter 9. His center of mass is at elevation 1.02 m when he leaves the floor. It reaches a maximum height of 1.85 m above the floor and is at elevation 0.900 m when he touches down again. Determine (a) his time of flight (his “hang time”), (b) his horizontal and (c) vertical velocity components at the instant of takeoff, and (d) his takeoff angle. (e) For comparison, determine the hang time of a whitetail deer making a jump (Fig. P4.24b) with center-of-mass elevations yi 5 1.20 m, ymax 5 2.50 m, and yf 5 0.700 m.

b

Figure P4.24

B. G

. Sm

ith/S

hutt

erst

ock.

com

a

© e

pa e

urop

ean

pres

spho

to a

genc

y b.v.

/ Al

amy

25. A playground is on the flat roof of a city school, 6.00 m above the street below (Fig. P4.25). The vertical wall of the building is h 5 7.00 m high, forming a 1-m-high railing around the playground. A ball has fallen to the street below, and a passerby returns it by launch-ing it at an angle of u 5 53.0° above the horizontal at a point d 5 24.0 m from the base of the building wall. The ball takes 2.20 s to reach a point vertically above the wall. (a) Find the speed at which the ball was launched. (b) Find the vertical distance by which the ball clears the wall. (c) Find the horizontal distance from the wall to the point on the roof where the ball lands.

h

u

d

Figure P4.25

26. The motion of a human body through space can be modeled as the motion of a particle at the body’s cen-ter of mass as we will study in Chapter 9. The compo-nents of the displacement of an athlete’s center of mass from the beginning to the end of a certain jump are described by the equations

xf 5 0 1 (11.2 m/s)(cos 18.5°)t

0.360 m 5 0.840 m 1 111.2 m/s 2 1sin 18.582t 2 12 19.80 m/s2 2t 2

where t is in seconds and is the time at which the ath-lete ends the jump. Identify (a) the athlete’s position and (b) his vector velocity at the takeoff point. (c) How far did he jump?

27. A soccer player kicks a rock horizontally off a 40.0-m-high cliff into a pool of water. If the player W

the same speed but at the optimal angle for maximum range? (c) What If? Would your answer to part (a) be different if the rock is thrown with the same speed on a different planet? Explain.

19. The speed of a projectile when it reaches its maximum height is one-half its speed when it is at half its maxi-mum height. What is the initial projection angle of the projectile?

20. A ball is tossed from an upper-story window of a build-ing. The ball is given an initial velocity of 8.00 m/s at an angle of 20.0° below the horizontal. It strikes the ground 3.00 s later. (a) How far horizontally from the base of the building does the ball strike the ground? (b) Find the height from which the ball was thrown. (c) How long does it take the ball to reach a point 10.0 m below the level of launching?

21. A firefighter, a distance d from a burning building, directs a stream of water from a fire hose at angle ui above the horizontal as shown in Figure P4.21. If the initial speed of the stream is vi, at what height h does the water strike the building?

d

h

i

ui

vS

Figure P4.21

22. A landscape architect is planning an artificial water-fall in a city park. Water flowing at 1.70 m/s will leave the end of a horizon-tal channel at the top of a vertical wall h 5 2.35 m high, and from there it will fall into a pool (Fig. P4.22). (a) Will the space behind the waterfall be wide enough for a pedestrian walkway? (b) To sell her plan to the city council, the architect wants to build a model to standard scale, which is one-twelfth actual size. How fast should the water flow in the channel in the model?

23. A placekicker must kick a football from a point 36.0 m (about 40 yards) from the goal. Half the crowd hopes the ball will clear the crossbar, which is 3.05 m high. When kicked, the ball leaves the ground with a speed of 20.0 m/s at an angle of 53.0° to the horizontal. (a) By how much does the ball clear or fall short of clearing the crossbar? (b) Does the ball approach the crossbar while still rising or while falling?

24. A basketball star covers 2.80 m horizontally in a jump to dunk the ball (Fig. P4.24a). His motion through space can be modeled precisely as that of a particle at his center

W

S

h

Figure P4.22

MAMT

1Serway & Jewett

Projectile Motion Example 4.5A ski jumper leaves the ski track moving in the horizontal directionwith a speed vi as shown. The landing incline below her falls off atan angle φ. Find an expression for d , the distance down the inclinewhere she lands.

90 Chapter 4 Motion in Two Dimensions

Express the coordinates of the jumper as a function of time, using the particle under constant velocity model for x and the position equation from the particle under constant acceleration model for y:

(1) xf 5 vxi t

(2) yf 5 vyit 2 12gt 2

(3) d cos f 5 vxit

(4) 2d sin f 5 212gt 2

Solve Equation (3) for t and substitute the result into Equation (4):

2d sin f 5 212g ad cos f

vxib2

Solve for d and substitute numerical values: d 52vxi

2 sin fg cos2 f

52 125.0 m/s 22 sin 35.0819.80 m/s2 2 cos2 35.08

5 109 m

Evaluate the x and y coordinates of the point at which the skier lands:

xf 5 d cos f 5 1109 m 2 cos 35.08 5 89.3 m

yf 5 2d sin f 5 2 1109 m 2 sin 35.08 5 262.5 m

Finalize Let us compare these results with our expectations. We expected the horizontal distance to be on the order of 100 m, and our result of 89.3 m is indeed on this order of magnitude. It might be useful to calculate the time interval that the jumper is in the air and compare it with our estimate of about 4 s.

Suppose everything in this example is the same except the ski jump is curved so that the jumper is pro-jected upward at an angle from the end of the track. Is this design better in terms of maximizing the length of the jump?

Answer If the initial velocity has an upward component, the skier will be in the air longer and should therefore travel farther. Tilting the initial velocity vector upward, however, will reduce the horizontal component of the initial veloc-ity. Therefore, angling the end of the ski track upward at a large angle may actually reduce the distance. Consider the extreme case: the skier is projected at 90° to the horizontal and simply goes up and comes back down at the end of the ski track! This argument suggests that there must be an optimal angle between 0° and 90° that represents a balance between making the flight time longer and the horizontal velocity component smaller. Let us find this optimal angle mathematically. We modify Equations (1) through (4) in the following way, assum-ing the skier is projected at an angle u with respect to the horizontal over a landing incline sloped with an arbitrary angle f:

(1) and (3) S xf 5 (vi cos u)t 5 d cos f

(2) and (4) S yf 5 (vi sin u)t 2 12gt 2 5 2d sin f

WHAT IF?

Example 4.5 The End of the Ski Jump

A ski jumper leaves the ski track moving in the horizontal direction with a speed of 25.0 m/s as shown in Figure 4.14. The landing incline below her falls off with a slope of 35.0°. Where does she land on the incline?

Conceptualize We can conceptualize this problem based on memories of observing winter Olympic ski competitions. We estimate the skier to be airborne for perhaps 4 s and to travel a distance of about 100 m hori-zontally. We should expect the value of d, the distance traveled along the incline, to be of the same order of magnitude.

Categorize We categorize the problem as one of a particle in projectile motion. As with other projectile motion problems, we use the particle under constant velocity model for the horizontal motion and the particle under constant acceleration model for the vertical motion.

Analyze It is convenient to select the beginning of the jump as the ori-gin. The initial velocity components are vxi 5 25.0 m/s and vyi 5 0. From the right triangle in Figure 4.14, we see that the jumper’s x and y coordi-nates at the landing point are given by xf 5 d cos f and yf 5 2d sin f.

AM

S O L U T IO N

y d

25.0 m/s

O

x

f ! 35.0"

Figure 4.14 (Example 4.5) A ski jumper leaves the track moving in a horizontal direction.

vi

1Serway & Jewett

Relative Motion

One very useful technique for physical reasoning is consideringother frames of reference.

A reference frame is a coordinate system that an observer adopts.

Different observers may have different axes: different frames ofreference.

Reference Frames

Two observers could agree to choose directions as North (y) andEast (x). However, they might pick different origins, O and O ′, fortheir axes.

In this case, each person would describe the location of a particleslightly differently. How can we relate those descriptions?

1Image modified from work of Wikipedia user Krea.

Reference FramesObservers A and B have different descriptions of the location ofparticle P. We can relate these descriptions using vector addition.

4.6 Relative Velocity and Relative Acceleration 97

will be separated by a distance vBAt. We label the position P of the particle relative to observer A with the position vector rSP A and that relative to observer B with the position vector rSP B, both at time t. From Figure 4.20, we see that the vectors rSP A and rSP B are related to each other through the expression

rSP A 5 rSP B 1 vSBAt (4.22)

By differentiating Equation 4.22 with respect to time, noting that vSBA is con-stant, we obtain

d rSP A

dt5

d rSP B

dt1 vSBA

uSP A 5 uSP B 1 vSBA (4.23)

where uSPA is the velocity of the particle at P measured by observer A and uSP B is its velocity measured by B. (We use the symbol uS for particle velocity rather than vS, which we have already used for the relative velocity of two reference frames.) Equa-tions 4.22 and 4.23 are known as Galilean transformation equations. They relate the position and velocity of a particle as measured by observers in relative motion. Notice the pattern of the subscripts in Equation 4.23. When relative velocities are added, the inner subscripts (B) are the same and the outer ones (P, A) match the subscripts on the velocity on the left of the equation. Although observers in two frames measure different velocities for the particle, they measure the same acceleration when vSBA is constant. We can verify that by taking the time derivative of Equation 4.23:

d uSP A

dt5

d uSP B

dt1

d vSBA

dt

Because vSBA is constant, d vSBA/dt 5 0 . Therefore, we conclude that aSP A 5 aSP B because aSP A 5 d uSP A/dt and aSP B 5 d uSP B/dt. That is, the acceleration of the parti-cle measured by an observer in one frame of reference is the same as that measured by any other observer moving with constant velocity relative to the first frame.

WW Galilean velocity transformation

continued

Example 4.8 A Boat Crossing a River

A boat crossing a wide river moves with a speed of 10.0 km/h relative to the water. The water in the river has a uniform speed of 5.00 km/h due east relative to the Earth.

(A) If the boat heads due north, determine the velocity of the boat relative to an observer standing on either bank.

Conceptualize Imagine moving in a boat across a river while the current pushes you down the river. You will not be able to move directly across the river, but will end up downstream as suggested in Figure 4.21a.

Categorize Because of the combined velocities of you rela-tive to the river and the river relative to the Earth, we can categorize this problem as one involving relative velocities.

Analyze We know vSbr, the velocity of the boat relative to the river, and vSrE, the velocity of the river relative to the Earth. What we must find is vSbE, the velocity of the boat relative to the Earth. The relationship between these three quantities is vSbE 5 vSbr 1 vSrE. The terms in the equation must be manipulated as vector quantities; the vectors are shown in Fig-ure 4.21a. The quantity vSbr is due north; vSrE is due east; and the vector sum of the two, vSbE, is at an angle u as defined in Figure 4.21a.

S O L U T IO N u

br

bE

rE

E

N

S

W

a

vS

vS

vS

E

N

S

W

b

ubr

bE

rEvS

vS

vS

Figure 4.21 (Example 4.8) (a) A boat aims directly across a river and ends up downstream. (b) To move directly across the river, the boat must aim upstream.

SA SB

BA

P

xBAt

BA

PB

PA

vS vS

rS rS

Figure 4.20 A particle located at P is described by two observers, one in the fixed frame of refer-ence SA and the other in the frame SB, which moves to the right with a constant velocity vSBA. The vector rSPA is the particle’s position vector relative to SA, and rSP B is its position vector relative to SB.

r

#»rPA = #»rPB + #»rBA

#»rPA is the position of particle P relative to frame A.

Relative Motion

We can use the notion of motion in 2 dimensions to consider howone object moves relative to something else.

All motion is relative.

Our reference frame tells us what is a fixed position.

An example of a reference might be picking an object, declaringthat it is at rest, and describing the motion of all objects relativeto that.

Two different people could pick different reference objects and endup with two reference frames moving relative to each other.

Frames of Reference

Now consider observer B moving with constant velocity #»vBArelative to A. Both observe a particle P.





d rSP A

dt5

d rSP B

dt1 vSBA



d uSP A

dt5

d uSP B

dt1

d vSBA

dt

Because vSBA is constant, d vSBA/dt 5 0. Therefore, we conclude that aSP A 5 aSP B because aSP A 5 d uSP A/dt and aSP B 5 d uSP B/dt. That is, the acceleration of the parti-cle measured by an observer in one frame of reference is the same as that measured by any other observer moving with constant velocity relative to the first frame.


continued







S O L U T I O N u

br

bE

rE

E

N

S

W

a

vS

vS

vS

E

N

S

W

b

ubr

bE

rEvS

vS

vS


SA SB

BA

P

xBAt

BA

PB

PA

vS vS

rS rS


(In this diagram, the frames coincide at t = 0.)

Relative Motion





d rSP A

dt5

d rSP B

dt1 vSBA



d uSP A

dt5

d uSP B

dt1

d vSBA

dt

Because vSBA is constant, d vSBA/dt 5 0. Therefore, we conclude that aSP A 5 aSP B because aSP A 5 d uSP A/dt and aSP B 5 d uSP B/dt. That is, the acceleration of the parti-cle measured by an observer in one frame of reference is the same as that measured by any other observer moving with constant velocity relative to the first frame.


continued







S O L U T I O N u

br

bE

rE

E

N

S

W

a

vS

vS

vS

E

N

S

W

b

ubr

bE

rEvS

vS

vS


SA SB

BA

P

xBAt

BA

PB

PA

vS vS

rS rS


#»rPA = #»rPB + #»vBAt

Differentiating, noting that #»v BA is a constant:

d #»rPA

dt=

d #»rPB

dt+ #»vBA

#»vPA = #»vPB + #»vBA

Differentiating again shows that observers in two frames related bya constant relative velocity see the same accelerations.

Summary

• projectiles

• introduced relative motion (?)

Quiz start of class tomorrow (Friday).

(Uncollected) Homework Serway & Jewett,

• Ch 4, onward from page 104. Problems: 59

2d kinematics projectiles relative...

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