CHAPTER 3
2.
REASONING AND SOLUTION An object thrown upward at an angle  will
follow the trajectory shown below. Its acceleration is that due to gravity, and,
therefore, always points downward. The acceleration is denoted by ay in the figure.
In general, the velocity of the object has two components, vx and vy. Since ax = 0,
vx always equals its initial value. The y component of the velocity, vy, decreases as
the object rises, drops to zero when the object is at its highest point, and then
increases in magnitude as the object falls downward.
a. Since vy = 0 when the object is at its highest point, the velocity of the object
points only in the x direction. As suggested in the figure below, the acceleration will
be perpendicular to the velocity when the object is at its highest point and vy = 0.
ay
ay
v0y
v0x
ay

v0x
v0x

–v0y
vf
b. In order for the velocity and acceleration to be parallel, the x component of the
velocity would have to drop to zero. However, vx always remains equal to its initial
value; therefore, the velocity and the acceleration can never be parallel.
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6.
REASONING AND SOLUTION If there were a wind blowing parallel to the
ground and toward the kicker in Figure 3.12, the acceleration component in the
horizontal direction would not be zero. The horizontal component of the velocity
would decrease as time goes on. The flight time of the ball, however, depends only
on the vertical component of the initial velocity. Thus, the flight time of the ball
would not be affected by the wind. The horizontal range of the football will be
shortened, but the flight time will remain the same.
10. REASONING AND SOLUTION
a. The displacement is greater for the stone that is thrown horizontally, because it
has the same vertical component as the dropped stone and, in addition, has a
horizontal component.
b. The impact speed is greater for the stone that is thrown horizontally. The reason
is that it has the same vertical velocity component as the dropped stone but, in
addition, also has a horizontal component that equals the throwing velocity.
c. The time of flight is the same in each case, because the vertical part of the motion
for each stone is the same. That is, each stone has an initial vertical velocity
component of zero and falls through the same height.
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15. REASONING AND SOLUTION
a. Since the child is at rest with respect to the floor of the RV and with respect to the
cannon, there is no relative motion between the child, the floor of the RV and the
cannon. Therefore, the range of the marble toward the front is the same as the range
toward the rear from the point of view of the child.
b. The velocity of the marble relative to an observer standing still on the ground is
v MG  v MV  v VG , where v MV is the velocity of the marble relative to the vehicle
and v VG is the velocity of the vehicle relative to the ground. Suppose that the RV
moves in the positive direction; then the quantity v VG is positive. When the
marble is shot toward the front of the RV, the x component of v MV will also be a
positive quantity. Alternatively, when the marble is shot toward the back of the RV,
the x component of v MV is a negative quantity. Thus, the magnitude of the x
component of v MG is smaller when the marble is shot toward the rear than when it is
shot toward the front. The time that the marble is in the air is determined by the
vertical motion which is the same in both cases. Since the magnitude of the x
component of v MG is smaller when the marble is shot toward the rear, it will cover
less distance in the time it takes to hit the floor of the RV compared to when the
marble is shot toward the front. Therefore, from the point of view of an observer on
the ground, the range of the marble is greater when the marble is shot toward the
front.
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9.
SSM WWW
REASONING AND SOLUTION The
escalator is the hypotenuse of a right triangle formed by the
lower floor and the vertical distance between floors, as shown
in the figure. The angle  at which the escalator is inclined
above the horizontal is related to the length L of the escalator
and the vertical distance between the floors by the sine
function:
6.00 m
sin  
L
(1)
The length L of the escalator can be found from the right triangle formed by the
components of the shopper's displacement up the escalator and to the right of the
escalator.
From the Pythagorean theorem, we have
9.00 m
2
2
L  (9.00 m)  (16.0 m)
2
so that
L
16.0 m
L  (16.0 m) 2  ( 9.00 m) 2  13.2 m
From Equation (1) we have
6.00 m 
 27.0
13.2 m 
  sin 1 
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13. REASONING AND SOLUTION The vertical motion consists of the ball rising for a
time, t, stopping and returning to the ground in another time, t. For the upward
portion
t =
voy - vy
voy
=
g
g
Note: vy = 0 since the ball stops at the top. Now
voy = vo sin o = (25.0 m/s) sin 60.0° = 21.7 m/s
t = (21.7 m/s)/(9.80 m/s2) = 2.21 s
The required "hang time" is 2t = 4.42 s .
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16. REASONING AND SOLUTION Using vy = 0 and
voy = vo sin  = (11 m/s) sin 65 = 1.0  101 m/s
and vy2 = voy2 + 2ayy, we have
y =
- v2oy
2a y
1
=
2
- (1.0 X 10 m/s)
2(- 9.80 m/s2)
= 5.1 m
26. REASONING AND SOLUTION The maximum vertical displacement y attained
2
by a projectile is given by Equation 3.6b ( v 2y  v0y
 2ay y ) with v y  0 :
y–
2
v0y
2a y
In order to use Equation 3.6b, we must first estimate his initial speed v0y . When
Jordan has reached his maximum vertical displacement, v y  0 , and t  1.00 s .
Therefore, according to Equation 3.3b ( v y  v0y  ayt ), with upward taken as
positive, we find that
v 0 y  –a y t  – (–9.80 m/s 2 ) (1.00 s) = 9.80 m/s
Therefore, Jordan's maximum jump height is
y–
(9.80 m/s)2
2(–9.80 m/s2 )
 4.90 m
This result far exceeds Jordan’s maximum jump height, so the claim that he can
remain in the air for two full seconds is false.
27. REASONING AND SOLUTION The time of flight of the motorcycle is given by
o
2vo sin  o
2(33.5 m/s) sin 18.0
t=
=
= 2.11 s.
g
9.80 m/s2
The horizontal distance traveled by the motorcycle is then
x = vo cos o t = (33.5 m/s)(cos18.0°)(2.11 s) = 67.2 m.
The daredevil can jump over (67.2 m)/(2.74 m/bus) = 24.5 buses. In even numbers,
this means 24 buses .
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30. REASONING The rocket will clear the top of the wall by an amount that is the
height of the rocket as it passes over the wall minus the height of the wall. To find
the height of the rocket as it passes over the wall, we separate the rocket’s projectile
motion into its horizontal and vertical parts and treat each one separately. From the
horizontal part we will obtain the time of flight until the rocket reaches the location
of the wall. Then, we will use this time along with the acceleration due to gravity in
the equations of kinematics to determine the height of the rocket as it passes over the
wall.
SOLUTION We begin by finding the horizontal and vertical components of the
launch velocity
v 0 x  v 0 cos 60.0   75.0 m / s cos 60.0 
v0 y  v0
b
g
sin 60.0   b
75.0 m / s g
sin 60.0 
Using v0x, we can obtain the time of flight, since the distance to the wall is known to
be 27.0 m:
27 .0 m
27 .0 m
t

 0.720 s
v0 x
75.0 m / s cos 60.0
b
g
The height of the rocket as it clears the wall can be obtained from Equation 3.5b, in
which we take upward to be the positive direction. The amount by which the rocket
clears the wall can then be obtained:
y  v 0 y t  21 a y t 2
b
gb
gb
gc
y  75.0 m / s sin 60.0  0.720 s  21 9 .80 m / s 2
hb0.720 sg 44.2 m
2
clearance  44 .2 m  11.0 m = 33.2 m
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36. REASONING AND SOLUTION On impact
vx = v cos 75.0° = (8.90 m/s) cos 75.0° = 2.30 m/s
and
v0y2 = vy2 + 2gy = (8.90 m/s)2 sin2 75.0° + 2(9.80 m/s2)(–3.00 m)
so
v0y = 3.89 m/s.
The magnitude of the diver's initial velocity is
vo =
2
2
(2.30 m/s) + (3.89 m/s) = 4.52 m/s
The angle the initial velocity vector makes with the horizontal is
0 = tan-1 (voy /vox) =
59.4
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39. REASONING AND SOLUTION The horizontal position of the impact point is x1 =
v0xt1 for one stone and x2 = v0xt2 for the other stone.
The vertical positions are y1 = – (1/2)gt12 and y2 = – (1/2)gt22. Dividing the x
equations gives
x2/x1 = t2/t1 = 2
Dividing the y equations gives
y2/y1 = t22/t12 = 4
So one building is four times as tall as the other.
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49. REASONING AND SOLUTION The speed of a person relative to the ground is
v = x/t = (105 m)/(75 s) = 1.4 m/s
If the person walks at this rate relative to the conveyor belt, his speed relative to the
ground is
v = 1.4 m/s + 2.0 m/s = 3.4 m/s
Thus,
t' = x/v' = (105 m)/(3.4 m/s) = 31 s
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51. SSM REASONING The velocity v SG of the
swimmer relative to the ground is the vector sum
of the velocity v SW of the swimmer relative to the
water and the velocity v WG of the water relative
to the ground as shown at the right:
v SG  v SW  vWG .
The component of v SG that is parallel to the
width of the river determines how fast the
swimmer is moving across the river; this parallel
component is vSW . The time for the swimmer to
cross the river is equal to the width of the river
divided by the magnitude of this velocity
component.
The component of v SG that is parallel to the direction of the current determines how
far the swimmer is carried down stream; this component is v WG . Since the motion
occurs with constant velocity, the distance that the swimmer is carried downstream
while crossing the river is equal to the magnitude of v WG multiplied by the time it
takes for the swimmer to cross the river.
SOLUTION
a. The time t for the swimmer to cross the river is
width 2.8  103 m
t

 2.0  103 s
vSW
1.4 m/s
b. The distance x that the swimmer is carried downstream while crossing the river is
x  v WGt  (0.91 m/s)(2.0 103 s) = 1.8 10 3 m
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