th
12 Two balls are launched upward from the same spot at different angles with respect to the ground. Both balls rise to the same maximum height. Ball A, however, follows a trajectory that has a greater range than that of ball B. Ignoring air resistance, decide which ball, if either, has the greater launch speed. Give your reasoning.
*34 See Multiple-Concept Example 9 for the basic idea behind problems such as this. A diver springs upward from a board that is three meters above the water. At the instant she contacts the water her speed is 8.90 m/s and her body makes an angle of 75.0° with respect to the horizontal surface of the water. Determine her initial velocity, both magnitude and direction.
**44 A baseball is hit into the air at an initial speed of 36.6 m/s and an angle of 50.0° above the horizontal. At the same time, the center fielder starts running away from the batter, and he catches the ball 0.914 m above the level at which it was hit. If the center fielder is initially 110m from home plate, what must be his average speed?
*56 Relative to the ground, a car has a velocity of 18.0 m/s, directed due north. Relative to this car, a truck has a velocity of 22.8 m/s, directed 52.1° south of east. Find the magnitude and direction of the truck’s velocity relative to the ground.
*8 An arrow, starting from rest, leaves the bow with a speed of 25.0 m/s. If the average force exerted on the arrow by the bow were doubled, all else remaining the same, with what speed would the arrow leave the bow?

12 . REASONING AND SOLUTION Since air resistance is being ignored, the equations of kinematics apply. Therefore, the maximum height is determined solely by the vertical component v
0 y
of the launch velocity and the acceleration due to gravity ( a y
=  9.80 m/s
2
, upward is assumed to be the positive direction). This can be seen from Equation 3.6b
 v
2 y
 v
2
0 y

2

, which can be solved for the vertical displacement y when v y
= 0 m/s at the maximum height of y = H : v
2 y
 v
2
0 y

2 a y
  2  v
2
0 y

2 a y or y

H

 v
0
2 y
2 a y
Since both balls attain the same maximum height, their launch velocities must have the same vertical components v
0 y
. In addition, the time of flight, which is the time for the balls to rise upward to the maximum height and fall back to earth, is the same for each ball. This can be seen from Equation 3.3b
 v y
 v
0 y
 a t y

, which can be solved for the time t when v y
= 0 m/s at the maximum height:
 v
0 y v y
 v
0 y
 a t or 0 m/s
 v
0 y
 a t or t
 a y
The time of flight is twice this value of t . Since v
0y
is the same for each ball, the time of flight is the same. But the range is the product of the horizontal component v
0 x
of the launch velocity and the time of flight. Thus, the greater range of ball A means that it must have a greater value for v
0 x
than does ball B, since the flight times are identical. Because both balls have the same value for v
0 y
and ball A has the greater value for v
0 x
, we conclude that ball A must have the greater launch speed v
0
, which is the magnitude of the launch velocity and is v
0
 v
2
0 x
 v
2
0 y
.
34.
REASONING AND SOLUTION On impact v x
= v cos 75.0° = (8.90 m/s) cos 75.0° = 2.30 m/s and v
0 y
2 = v y
2 + 2 gy = (8.90 m/s) 2 sin 2 75.0° + 2(9.80 m/s 2 )(–3.00 m) so v
0 y
= 3.89 m/s
The magnitude of the diver's initial velocity is v

0

2.30 m/s
 2  
3.89 m/s
 2 
4.52 m/s
The angle the initial velocity vector makes with the horizontal is

0
= tan

1 ( v o y
/ v o x
) =
59.4


44 . REASONING AND SOLUTION The ball is caught at a position of y = + 0.914 m and x = v
F t + 1.10  10 2 m where t is the time of flight and v
F
is the speed of the center fielder. v
F
 x
 t

 v
0 x

 t
The time t can be found from y = v o yt
– (1/2) gt 2 = ( v o sin
 o) t – (1/2) gt 2
or 0.914 m = (36.6 m/s)(sin 50.0°) t – (4.9 m/s 2 ) t 2
Rearrangement gives
4.90 t
2
– 28.0
t + 0.914 = 0, where the units have been suppressed.
The quadratic formula gives t = 5.68 s. Hence v
F
= (36.6 m/s) cos 50.0o – (1.10  10 2 m)/(5.68 s) =
4.2 m/s
56 . REASONING The velocities in this problem are as follows: v
CG v
TC
= the velocity of the C ar relative to the G round = 18.0 m/s, due north
= the velocity of the T ruck relative to the C ar = 22.8 m/s, 52.1
 south of east v
TG
= the velocity of the T ruck relative to the G round (unknown)
Using the ordering of the subscripts as discussed in the text, we can find the velocity of the truck relative to the ground by the following vector addition: v
TG
= v
TC
+ v
CG
North v
52.1

TC v
TG v
CG
The drawing (not to scale) illustrates this vector addition by arranging the vectors v
TC
and v method to carry out the addition.
CG
in tail-to-head fashion. We will use the component
SOLUTION In implementing the component method, we choose east and north to be the + x and + y directions, respectively. The x component of v
TG
is the sum of the x components of

v
TC v
CG
and v
CG
. Similarly, the
. Thus, we have y component of v
TG
is the sum of the y components of v
TC
and v
TG, x
 v
TC, x
 v
CG, x
     
  
18.0 m/s

0 m/s v
TG, y
 v
TC, y
 v
CG, y
  
Note that the component v
CG, x
is zero, since v
CG
points due north and has no component along the east-west line (the x axis). Note also that the component v
TC, y v
TC
is negative since
points below the east-west line. Using the Pythagorean theorem, we find that the magnitude of v
TG
is v
TG
 v
2
TG, x
 v
2
TG, y
 
14.0 m/s
 
0 m/s
 2 
14.0 m/s
Since v
TG, x
is positive and since v
TC, y
is zero, the direction of v
TG
is along the + x axis or
due east .
8.
REASONING AND SOLUTION From Equation 2.9, v
2  v
2
0

2 ax
Since the arrow starts from rest, v
0
= 0 m/s. In both cases x is the same so v
1
2

2 a x
 a
1
2 a x a
2
or v
1 v
2
 a
1 a
2
Since F = ma , it follows that a = F / m . The mass of the arrow is unchanged, and v
2
2 v
1 v
2

F
1
F
2 or v
2
 v
1
F
2
F
1
 v
1
2 F
1
F
1
 
25.0 m/s

2

35.4 m/s