Cutnell_9th_problems_ch_1_thru_10_converted

Answer:
positive for segments and , negative for segment , zero for segment

REASONING The slope of a straight-line segment in a position-versus-time graph is the average
velocity.

The algebraic sign of the average velocity, therefore, corresponds to the sign of the slope.

SOLUTION

(a) The slope, and hence the average velocity, is positive for segments and , negative for segment ,
and zero for segment .

(b) In the given position-versus-time graph, we find the slopes of the four straight-line segments to be
(b) Calculate the average velocity for each segment to verify your answers to part (a).

Answer:

for segment ,

for segment ,

for segment ,

for

segment

REASONING The slope of a straight-line segment in a position-versus-time graph is the average
velocity.

The algebraic sign of the average velocity, therefore, corresponds to the sign of the slope.

SOLUTION

(a) The slope, and hence the average velocity, is positive for segments and , negative for segment ,
and zero for segment .

(b) In the given position-versus-time graph, we find the slopes of the four straight-line segments to be

REASONING The slope of a straight-line segment in a position-versus-time graph is the average
velocity. The algebraic sign of the average velocity, therefore, corresponds to the sign of the slope.

SOLUTION

(a) The slope, and hence the average velocity, is positive for segments and , negative for segment ,
and zero for segment .

(b) In the given position-versus-time graph, we find the slopes of the four straight-line segments to be
66. Starting at

at time

, an object takes

to travel
in the
direction at a constant velocity.
Make a position–time graph of the object's motion and calculate its velocity.
67.
A snowmobile moves according to the velocity–time graph shown in the drawing. What is the
snowmobile's average acceleration during each of the segments , , and ?
Answer:
for segment ,
for segment ,
for segment
68. A bus makes a trip according to the position–time graph shown in the drawing. What is the
average velocity (magnitude and direction) of the bus during each of the segments , , and ? Express
your answers in km/h.

*69.
A bus makes a trip according to the position–time graph shown in the illustration. What is the average
acceleration (in
) of the bus for the entire 3.5-h period shown in the graph?
Answer:

*70. A runner is at the position

when time

. One hundred meters away is the finish line. Every ten seconds,

this runner runs half the remaining distance to the finish line. During each ten-second segment, the
runner has a constant velocity. For the first forty seconds of the motion, construct

(a) the position–time graph and

(b) the velocity–time graph.

**71.

Two runners start one hundred meters apart and run toward each other. Each runs ten meters during
the first second. During each second thereafter, each runner runs ninety percent of the distance he ran
in the previous second.

Thus, the velocity of each person changes from second to second. However, during any one second,
the velocity remains constant. Make a position–time graph for one of the runners. From this graph,
determine

(a) how much time passes before the runners collide and

Answer:

REASONING The two runners start one hundred meters apart and run toward each other. Each runs
ten meters during the first second and, during each second thereafter, each runner runs ninety percent
of the distance he ran in the previous second. While the velocity of each runner changes from second
to second, it remains constant during any one second.

SOLUTION The following table shows the distance covered during each second for one of the
runners, and the position at the end of each second (assuming that he begins at the origin) for the first
eight seconds.

T ime t (s) Dist ance co vered (m) Po sit io n x (m)

0.00

0.00

1.00

10.00

10.00

2.00

9.00
19.00
3.00
8.10
27.10
4.00
7.29
34.39
5.00
6.56
40.95
6.00
5.90
46.85
7.00
5.31
52.16
8.00
4.78
56.94
The following graph is the position-time graph constructed from the data in the table above.

(a) Since the two runners are running toward each other in exactly the same way, they will meet
halfway between their respective starting points. That is, they will meet at

. According to the graph,

therefore, this position corresponds to a time of

.

(b) Since the runners collide during the seventh second, the speed at the instant of collision can be
found by taking the slope of the position-time graph for the seventh second. The speed of either
runner in the interval from = 6.00 s to = 7.00 s is

Therefore, at the moment of collision, the speed of either runner is

.

(b) the speed with which each is running at the moment of collision.

Answer:

REASONING The two runners start one hundred meters apart and run toward each other. Each runs
ten meters during the first second and, during each second thereafter, each runner runs ninety percent
of the distance he ran in the previous second. While the velocity of each runner changes from second
to second, it remains constant during any one second.

SOLUTION The following table shows the distance covered during each second for one of the
runners, and the position at the end of each second (assuming that he begins at the origin) for the first
eight seconds.

T ime t (s) Dist ance co vered (m) Po sit io n x (m)

0.00

0.00

1.00

10.00

10.00
2.00
9.00
19.00
3.00
8.10
27.10
4.00
7.29
34.39
5.00
6.56
40.95
6.00
5.90
46.85
7.00
5.31
52.16
8.00
4.78
56.94

The following graph is the position-time graph constructed from the data in the table above.

(a) Since the two runners are running toward each other in exactly the same way, they will meet
halfway between their respective starting points. That is, they will meet at

. According to the graph,

therefore, this position corresponds to a time of

.

(b) Since the runners collide during the seventh second, the speed at the instant of collision can be
found by taking the slope of the position-time graph for the seventh second. The speed of either
runner in the interval from = 6.00 s to = 7.00 s is

Therefore, at the moment of collision, the speed of either runner is

.

REASONING The two runners start one hundred meters apart and run toward each other. Each runs
ten meters during the first second and, during each second thereafter, each runner runs ninety percent
of the distance he ran in the previous second. While the velocity of each runner changes from second
to second, it remains constant during any one second.

SOLUTION The following table shows the distance covered during each second for one of the
runners, and the position at the end of each second (assuming that he begins at the origin) for the first
eight seconds.

T ime t (s) Dist ance co vered (m) Po sit io n x (m)

0.00

0.00

1.00

10.00

10.00

2.00
9.00
19.00
3.00
8.10
27.10
4.00
7.29
34.39
5.00
6.56
40.95
6.00
5.90
46.85
7.00
5.31
52.16
8.00
4.78
56.94
The following graph is the position-time graph constructed from the data in the table above.

(a) Since the two runners are running toward each other in exactly the same way, they will meet
halfway between their respective starting points. That is, they will meet at

. According to the graph, therefore, this

position corresponds to a time of

.

(b) Since the runners collide during the seventh second, the speed at the instant of collision can be
found by taking the slope of the position-time graph for the seventh second. The speed of either
runner in the interval from =

6.00 s to = 7.00 s is

Therefore, at the moment of collision, the speed of either runner is

.

Copyright © 2012 John Wiley & Sons, Inc. Al rights reserved.

Problems

Section 3.1 Displacement, Velocity, and

Acceleration

1.

Two trees have perfectly straight trunks and are both growing perpendicular to the flat horizontal
ground beneath them. The sides of the trunks that face each other are separated by 1.3 m. A frisky

squirrel makes three jumps in rapid succession. First, he leaps from the foot of one tree to a spot that
is 1.0 m above the ground on the other tree. Then, he jumps back to the first tree, landing on it at a
spot that is 1.7 m above the ground. Finally, he leaps back to the other tree, now landing at a spot that
is 2.5 m above the ground. What is the magnitude of the squirrel's displacement?
Answer:
2.8 m
REASONING The displacement is a vector drawn from the initial position to the final position. The
magnitude of the displacement is the shortest distance between the positions. Note that it is only the
initial and final positions that determine the displacement. The fact that the squirrel jumps to an
intermediate position before reaching his final position is not important. The trees are perfectly
straight and both growing perpendicular to the flat horizontal ground beneath them. Thus, the distance
between the trees and the length of the trunk of the second tree below the squirrel's final landing spot
form the two perpendicular sides of a right triangle, as the drawing shows. To this triangle, we can
apply the Pythagorean theorem and determine the magnitude of the displacement vector A.

SOLUTION According to the Pythagorean theorem, we have
2. A meteoroid is traveling east through the atmosphere at 18.3 km/s while descending at a rate of 11.5

km/s. What is its speed, in km/s?

3.

In a football game a kicker attempts a field goal. The ball remains in contact with the kicker's foot for
0.050 s, during which time it experiences an acceleration of

. The ball is launched at an angle of

above the

ground. Determine the horizontal and vertical components of the launch velocity.

Answer:

4. A baseball player hits a triple and ends up on third base. A baseball “diamond” is a square, each
side of length 27.4 m, with home plate and the three bases on the four corners. What is the magnitude
of the player's displacement?

5.

In diving to a depth of 750 m, an elephant seal also moves 460 m due east of his starting point. What
is the magnitude of the seal's displacement?

Answer:

REASONING The displacement of the elephant seal has two components; 460 m due east and 750 m
downward. These components are mutually perpendicular; hence, the Pythagorean theorem can be
used to determine their resultant.

SOLUTION From the Pythagorean theorem,

Therefore,

6. A mountain-climbing expedition establishes two intermediate camps, labeled A and B in the
drawing, above the base camp. What is the magnitude

of the displacement between camp A and camp B?

7.

A radar antenna is tracking a satellite orbiting the earth. At a certain time, the radar screen shows the
satellite to be 162 km away. The radar antenna is pointing upward at an angle of

from the ground. Find the x and y

components (in km) of the position vector of the satellite, relative to the antenna.

Answer:

,

REASONING AND SOLUTION

8.

In a mall, a shopper rides up an escalator between floors. At the top of the escalator, the shopper turns
right and walks 9.00 m to a store. The magnitude of the shopper's displacement from the bottom of
the escalator to the store is 16.0 m. The vertical distance between the floors is 6.00 m. At what angle is
the escalator inclined above the horizontal?

9.

A skateboarder, starting from rest, rolls down a 12.0-m ramp. When she arrives at the bottom of the
ramp her speed is

.

(a) Determine the magnitude of her acceleration, assumed to be constant.

Answer:

REASONING

(a) We designate the direction down and parallel to the ramp as the

direction, and the table shows the

variables that are known. Since three of the five kinematic variables have values, one of the equations
of kinematics can be employed to find the acceleration

.

x-Direction Data

?

0 m/s

(b) The acceleration vector points down and parallel to the ramp, and the angle of the ramp is relative
to the ground (see the drawing). Therefore, trigonometry can be used to determine the

component

of the acceleration that is parallel to the ground.
SOLUTION
(a)
Equation 3.6a
can be used to find the acceleration in terms of the three known
variables. Solving this equation for
gives

(b) The drawing shows that the acceleration vector is oriented 25.0 relative to the ground. The
component of the acceleration that is parallel to the ground is
(b) If the ramp is inclined at
with respect to the ground, what is the component of her acceleration that is
parallel to the ground?
Answer:
REASONING
(a) We designate the direction down and parallel to the ramp as the

direction, and the table shows the
variables that are known. Since three of the five kinematic variables have values, one of the equations
of kinematics can be employed to find the acceleration
.
x-Direction Data
?
0 m/s
(b) The acceleration vector points down and parallel to the ramp, and the angle of the ramp is relative
to the ground (see the drawing). Therefore, trigonometry can be used to determine the
component
of the acceleration that is parallel to the ground.
SOLUTION
(a)
Equation 3.6a
can be used to find the acceleration in terms of the three known
variables. Solving this equation for
gives
(b) The drawing shows that the acceleration vector is oriented 25.0 relative to the ground. The
component of the acceleration that is parallel to the ground is
REASONING
(a) We designate the direction down and parallel to the ramp as the
direction, and the table shows the variables
that are known. Since three of the five kinematic variables have values, one of the equations of
kinematics can be employed to find the acceleration
.
x-Direction Data
?

0 m/s
(b) The acceleration vector points down and parallel to the ramp, and the angle of the ramp is relative
to the
ground (see the drawing). Therefore, trigonometry can be used to determine the component
of the
acceleration that is parallel to the ground.

SOLUTION
(a)
Equation 3.6a
can be used to find the acceleration in terms of the three known variables.
Solving this equation for
gives
(b) The drawing shows that the acceleration vector is oriented 25.0 relative to the ground. The
component of the acceleration that is parallel to the ground is
*10.
A bird watcher meanders through the woods, walking 0.50 km due east, 0.75 km due south, and 2.15
km in a direction
north of west. The time required for this trip is 2.50 h. Determine the magnitude and direction
(relative to due west) of the bird watcher's
(a) displacement and
(b) average velocity. Use kilometers and hours for distance and time, respectively.
*11.
The earth moves around the sun in a nearly circular orbit of radius
. During the three

summer months (an elapsed time of
), the earth moves one-fourth of the distance around the sun.
(a) What is the average speed of the earth?
Answer:
(b) What is the magnitude of the average velocity of the earth during this period?
Answer:
Section 3.2
Equations
of
Kinematics
in Two
Dimensions
Section 3.3 Projectile Motion
12. A spacecraft is traveling with a velocity of
along the
direction. Two engines are turned on for a
time of 842 s. One engine gives the spacecraft an acceleration in the
direction of
, while the
other gives it an acceleration in the
direction of
. At the end of the firing, find
(a) and
(b) .
13.

A volleyball is spiked so that it has an initial velocity of

directed downward at an angle of

below

the horizontal. What is the horizontal component of the ball's velocity when the opposing player
fields the ball?

Answer:

REASONING The vertical component of the ball's velocity

changes as the ball approaches the opposing

player. It changes due to the acceleration of gravity. However, the horizontal component does not
change, assuming that air resistance can be neglected. Hence, the horizontal component of the ball's
velocity when the opposing player fields the ball is the same as it was initially.

SOLUTION Using trigonometry, we find that the horizontal component is

14. As a tennis ball is struck, it departs from the racket horizontally with a speed of
. The ball hits the court at a
horizontal distance of 19.6 m from the racket. How far above the court is the tennis ball when it leaves
the racket?
15. A skateboarder shoots off a ramp with a velocity of
, directed at an angle of
above the horizontal. The
end of the ramp is 1.2 m above the ground. Let the x axis be parallel to the ground, the

direction be vertically
upward, and take as the origin the point on the ground directly below the top of the ramp.
(a) How high above the ground is the highest point that the skateboarder reaches?
Answer:
2.8 m
(b) When the skateboarder reaches the highest point, how far is this point horizontally from the end
of the ramp?
Answer:
2.0 m
16.
A puck is moving on an air hockey table. Relative to an x, y coordinate system at time
, the x components
of the puck's initial velocity and acceleration are
and
. The y components of
the puck's initial velocity and acceleration are
and
. Find the magnitude and
direction of the puck's velocity at a time of
. Specify the direction relative to the
axis.
17.
A spider crawling across a table leaps onto a magazine blocking its path. The initial velocity of the
spider is at an angle of
above the table, and it lands on the magazine 0.0770 s after leaving the table. Ignore
air resistance. How thick is the magazine? Express your answer in millimeters.

Answer:
9.4 mm
REASONING Since the spider encounters no appreciable air resistance during its leap, it can be
treated as a projectile. The thickness of the magazine is equal to the spider's vertical displacement
during the leap. The relevant data are as follows (assuming upward to be the
direction):
y-Direction Data
?
0.0770 s
SOLUTION We will calculate the spider's vertical displacement directly from
(Equation 3.5b):
To express this result in millimeters, we use the fact that 1 m equals 1000 mm:
18. A horizontal rifle is fired at a bull's-eye. The muzzle speed of the bullet is
. The gun is pointed directly at
the center of the bull's-eye, but the bullet strikes the target 0.025 m below the center. What is the
horizontal distance between the end of the rifle and the bull's-eye?
19.
A golfer imparts a speed of
to a ball, and it travels the maximum possible distance before landing on
the green. The tee and the green are at the same elevation.
(a) How much time does the ball spend in the air?
Answer:
4.37 s
(b) What is the longest hole in one that the golfer can make, if the ball does not roll when it hits the
green?
Answer:
93.5 m

20.

A golfer hits a shot to a green that is elevated 3.0 m above the point where the ball is struck. The ball
leaves the club at a speed of

at an angle of

above the horizontal. It rises to its maximum height and then falls

down to the green. Ignoring air resistance, find the speed of the ball just before it lands.
21.
In the aerials competition in skiing, the competitors speed down a ramp that slopes sharply upward at
the end.
The sharp upward slope launches them into the air, where they perform acrobatic maneuvers. The end
of a launch ramp is directed
above the horizontal. With this launch angle, a skier attains a height of 13 m above the end of
the ramp. What is the skier's launch speed?
Answer:
REASONING When the skier leaves the ramp, she exhibits projectile motion. Since we know the
maximum height attained by the skier, we can find her launch speed
using Equation 3.6b,
, where
.
SOLUTION At the highest point in her trajectory,
. Solving Equation 3.6b for
we obtain, taking upward
as the positive direction,
22.
A space vehicle is coasting at a constant velocity of
in the

direction relative to a space station.

The pilot of the vehicle fires a RCS (reaction control system) thruster, which causes it to accelerate at
in the

direction. After 45.0 s, the pilot shuts off the RCS thruster. After the RCS thruster is turned off, find
(a) the magnitude and

(b) the direction of the vehicle's velocity relative to the space station. Express the direction as an angle
measured from the

direction.

23.

As preparation for this problem, review Conceptual Example 10. The drawing shows two planes each
about to drop an empty fuel tank. At the moment of release each plane has the same speed of

, and each tank is at the

same height of 2.00 km above the ground. Although the speeds are the same, the velocities are
different at the instant of release, because one plane is flying at an angle of

above the horizontal and the other is flying at an angle of

below the horizontal. Find the magnitude and direction of the velocity with which the fuel tank hits the
ground if it is from

(a) plane A and

Answer:

,

with respect to the horizontal

REASONING Since the magnitude of the velocity of the fuel tank is given by

, it is necessary to know the velocity components

and

just before

impact. At the instant of release, the empty fuel tank has the same velocity as that of the

plane. Therefore, the magnitudes of the initial velocity components of the fuel tank are



given by
and
, where is the speed of the plane at the
instant of release. Since the motion has zero acceleration, the component of the
velocity of the plane remains equal to
for all later times while the tank is airborne.
The component of the velocity of the tank after it has undergone a vertical
displacement is given by Equation 3.6b.
SOLUTION
(a Taking up as the positive direction, the velocity components of the fuel tank just
) before it hits the ground are
From Equation 3.6b, we have
Therefore, the magnitude of the velocity of the fuel tank just before impact is
The velocity vector just before impact is inclined at an angle with respect to the
horizontal. This angle is
(b As shown in Conceptual Example 10, once the fuel tank in part a rises and falls to
) the same altitude at which it was released, its motion is identical to the fuel tank in part b. Therefore,
the velocity of the fuel tank in part b just before impact is
.
(b) plane B. In each part, give the directional angles with respect to the horizontal.
Answer:
,
with respect to the horizontal
REASONING Since the magnitude of the velocity of the fuel tank is given by

, it is
necessary to know the velocity components
and
just before impact. At the instant of release, the empty
fuel tank has the same velocity as that of the plane. Therefore, the magnitudes of the initial velocity
components of the fuel tank are given by
and
, where is the speed of the plane
at the instant of release. Since the motion has zero acceleration, the component of the velocity of the
plane remains equal to
for all later times while the tank is airborne. The component of the velocity of the tank
after it has undergone a vertical displacement is given by Equation 3.6b.
SOLUTION
(a) Taking up as the positive direction, the velocity components of the fuel tank just before it hits the
ground are
From Equation 3.6b, we have
Therefore, the magnitude of the velocity of the fuel tank just before impact is

The velocity vector just before impact is inclined at an angle with respect to the horizontal. This angle
is

(b) As shown in Conceptual Example 10, once the fuel tank in part a rises and falls to the same
altitude at which it was released, its motion is identical to the fuel tank in part b. Therefore, the
velocity of the fuel tank in part b just before impact is

.

REASONING Since the magnitude of the velocity of the fuel tank is given by

, it is necessary

to know the velocity components

and

just before impact. At the instant of release, the empty fuel tank has the

same velocity as that of the plane. Therefore, the magnitudes of the initial velocity components of the
fuel tank are given by

and

, where is the speed of the plane at the instant of release. Since the

motion has zero acceleration, the component of the velocity of the plane remains equal to

for all later times while

the tank is airborne. The component of the velocity of the tank after it has undergone a vertical
displacement is given by Equation 3.6b.

SOLUTION

(a) Taking up as the positive direction, the velocity components of the fuel tank just before it hits the
ground are From Equation 3.6b, we have

Therefore, the magnitude of the velocity of the fuel tank just before impact is

The velocity vector just before impact is inclined at an angle with respect to the horizontal. This angle
is (b) As shown in Conceptual Example 10, once the fuel tank in part a rises and falls to the same
altitude at which it was released, its motion is identical to the fuel tank in part b. Therefore, the
velocity of the fuel tank in part b just before impact is

.

24.

A criminal is escaping across a rooftop and runs off the roof horizontally at a speed of

, hoping to land

on the roof of an adjacent building. Air resistance is negligible. The horizontal distance between the
two buildings is D, and the roof of the adjacent building is 2.0 m below the jumping-off point. Find
the maximum value for D.

25. On a spacecraft, two engines are turned on for 684 s at a moment when the velocity of the craft
has x and y components of

and

. While the engines are firing, the craft undergoes a

displacement that has components of
and
. Find the x and y components of the
craft's acceleration.
Answer:

26.

In the absence of air resistance, a projectile is launched from and returns to ground level. It follows a
trajectory similar to that shown in Figure 3.10 and has a range of 23 m. Suppose the launch speed is
doubled, and the projectile is fired at the same angle above the ground. What is the new range?

27.

A fire hose ejects a stream of water at an angle of

above the horizontal. The water leaves the nozzle with

a speed of

. Assuming that the water behaves like a projectile, how far from a building should the fire hose be
located to hit the highest possible fire?

Answer:

30.0 m

REASONING AND SOLUTION The water exhibits projectile motion. The component of the motion
has zero acceleration while the component is subject to the acceleration due to gravity. In order to
reach the highest possible fire, the displacement of the hose from the building is , where, according to

Equation 3.5a (with

),

with equal to the time required for the water the reach its maximum vertical displacement. The time
can be found by considering the vertical motion. From Equation 3.3b,

When the water has reached its maximum vertical displacement,

. Taking up and to the right as the

positive directions, we find that

and

Therefore, we have

28. Baseball player A bunts the ball by hitting it in such a way that it acquires an initial velocity of
parallel to

the ground. Upon contact with the bat the ball is 1.2 m above the ground. Player B wishes to duplicate
this bunt, in so far as he also wants to give the ball a velocity parallel to the ground and have his ball
travel the same horizontal distance as player A's ball does. However, player B hits the ball when it is
1.5 m above the ground. What is the magnitude of the initial velocity that player B's ball must be
given?

29. A major-league pitcher can throw a baseball in excess of

. If a ball is thrown horizontally at this speed, how

much will it drop by the time it reaches a catcher who is 17.0 m away from the point of release?

Answer:

0.844 m

30. A quarterback claims that he can throw the football a horizontal distance of 183 m (200 yd).
Furthermore, he claims that he can do this by launching the ball at the relatively low angle of

above the horizontal. To evaluate this

claim, determine the speed with which this quarterback must throw the ball. Assume that the ball is
launched and caught at the same vertical level and that air resistance can be ignored. For comparison,
a baseball pitcher who can accurately throw a fastball at

(100 mph) would be considered exceptional.

31.

An eagle is flying horizontally at
with a fish in its claws. It accidentally drops the fish.
(a) How much time passes before the fish's speed doubles?
Answer:
1.1 s
REASONING The speed of the fish at any time is given by
, where
and
are
the and components of the velocity at that instant. Since the horizontal motion of the fish has zero
acceleration,
for all times . Since the fish is dropped by the eagle,
is equal to the horizontal
speed of the eagle and
. The component of the velocity of the fish for any time is given by



Equation 3.3b with
. Thus, the speed at any time t is given by
.
SOLUTION
(a) The initial speed of the fish is
. When the fish's speed
doubles,
. Therefore,
Assuming that downward is positive and solving for , we have
(b) When the fish's speed doubles again,
. Therefore,
Solving for , we have
Therefore, the additional time for the speed to double again is
.
(b) How much additional time would be required for the fish's speed to double again?
Answer:
1.3 s
REASONING The speed of the fish at any time is given by
, where

and
are
the and components of the velocity at that instant. Since the horizontal motion of the fish has zero
acceleration,
for all times . Since the fish is dropped by the eagle,
is equal to the horizontal
speed of the eagle and
. The component of the velocity of the fish for any time is given by
Equation 3.3b with
. Thus, the speed at any time t is given by
.
SOLUTION
(a) The initial speed of the fish is
. When the fish's speed
doubles,
. Therefore,
Assuming that downward is positive and solving for , we have
(b) When the fish's speed doubles again,
. Therefore,
Solving for , we have
Therefore, the additional time for the speed to double again is
.
REASONING The speed of the fish at any time is given by
, where
and
are the and



components of the velocity at that instant. Since the horizontal motion of the fish has zero
acceleration, for
all times . Since the fish is dropped by the eagle,
is equal to the horizontal speed of the eagle and
. The
component of the velocity of the fish for any time is given by Equation 3.3b with
. Thus, the speed at any
time t is given by
.
SOLUTION
(a) The initial speed of the fish is
. When the fish's speed doubles,
. Therefore,
Assuming that downward is positive and solving for , we have
(b) When the fish's speed doubles again,
. Therefore,
Solving for , we have
Therefore, the additional time for the speed to double again is
.

32.

The perspective provided by Multiple-Concept Example 9 is useful here. The highest barrier that a
projectile can clear is 13.5 m, when the projectile is launched at an angle of

above the horizontal. What is the projectile's

launch speed?

33. Consult Multiple-Concept Example 4 for background before beginning this problem. Suppose the
water at the top of Niagara Falls has a horizontal speed of

just before it cascades over the edge of the falls. At what vertical

distance below the edge does the velocity vector of the water point downward at a

angle below the horizontal?

Answer:

5.2 m

34.

On a distant planet, golf is just as popular as it is on earth. A golfer tees off and drives the ball 3.5
times as far as he would have on earth, given the same initial velocities on both planets. The ball is
launched at a speed of at an angle of

above the horizontal. When the ball lands, it is at the same level as the tee. On the distant

planet, what are

(a) the maximum height and

(b) the range of the ball?

35. A rocket is fired at a speed of

from ground level, at an angle of

above the horizontal. The rocket is

fired toward an 11.0-m-high wall, which is located 27.0 m away. The rocket attains its launch speed in
a negligibly short period of time, after which its engines shut down and the rocket coasts. By how
much does the rocket clear the top of the wall?

Answer:

33.2 m

36. A rifle is used to shoot twice at a target, using identical cartridges. The first time, the rifle is
aimed parallel to the ground and directly at the center of the bull's-eye. The bullet strikes the target at
a distance of below the center,

however. The second time, the rifle is similarly aimed, but from twice the distance from the target.
This time the bullet strikes the target at a distance of

below the center. Find the ratio

.

*37.

An airplane with a speed of

is climbing upward at an angle of

with respect to the horizontal.

When the plane's altitude is 732 m, the pilot releases a package.

(a) Calculate the distance along the ground, measured from a point directly beneath the point of
release, to where the package hits the earth.

Answer:



1380 m

REASONING
(a) The drawing shows the initial velocity
of the package when it is released. The initial speed of the
package is
. The component of its displacement along the ground is labeled as . The data for
the x direction are indicated in the data table below.
x-Direction Data
?
Since only two variables are known, it is not possible to determine from the data in this table. A value
for a third variable is needed. We know that the time of flight is the same for both the and motions, so
let's now look at the data in the direction.
y-Direction Data
?
Note that the displacement of the package points from its initial position toward the ground, so its
value is negative, i.e.,
. The data in this table, along with the appropriate equation of
kinematics, can be used to find the time of flight . This value for can, in turn, be used in conjunction
with the
direction data to determine .
(b) The drawing at the right shows the velocity of the package just before impact. The angle that the
velocity makes with respect to the ground can be found from the inverse tangent function as
. Once the time has been found in part (a), the values of
and
can be
determined from the data in the tables and the appropriate equations of kinematics.

SOLUTION

(a
) To determine the time that the package is in the air, we will use Equation 3.5b
and the data in the y-direction data table. Solving this quadratic equation for the time yields
We discard the first solution, since it is a negative value and, hence, unrealistic. The displacement can
be found using
, the data in the -direction data table, and Equation 3.5a:
(b The angle that the velocity of the package makes with respect to the ground is given by
)
. Since there is no acceleration in the direction
,
is the same
as
, so that
. Equation 3.3b can be employed with the -direction data to
find
:
Therefore,
where the minus sign indicates that the angle is
.
(b) Relative to the ground, determine the angle of the velocity vector of the package just before
impact.
Answer:
below the horizontal
REASONING
(a) The drawing shows the initial velocity
of the package when it is released. The initial speed of the

package is
. The component of its displacement along the ground is labeled as . The data for
the x direction are indicated in the data table below.

x-Direction Data

?

Since only two variables are known, it is not possible to determine from the data in this table. A value
for a third variable is needed. We know that the time of flight is the same for both the and motions, so
let's now look at the data in the direction.

y-Direction Data

?

Note that the displacement of the package points from its initial position toward the ground, so its
value is negative, i.e.,

. The data in this table, along with the appropriate equation of

kinematics, can be used to find the time of flight . This value for can, in turn, be used in conjunction
with the

direction data to determine .

(b) The drawing at the right shows the velocity of the package just before impact. The angle that the
velocity makes with respect to the ground can be found from the inverse tangent function as

. Once the time has been found in part (a), the values of