Download Chapter 4 solutions

Chapter 4 solutions
4.
Picture the Problem: The car moves up the 5.5° incline with constant acceleration, changing both its horizontal and vertical
displacement simultaneously.
Strategy: Find the magnitude of the displacement along the incline, and then independently find the horizontal and vertical
components of the displacement.
10.
Solution: 1. (a) Find the magnitude of the
displacement along the incline using equation 2-11:
r  0  v0t  12 at 2  0  0  12  2.0 m/s2  12 s   140 m
2. Find the horizontal component of r :
x  d cos   140 m  cos5.5  140 m
3. (b) Find the vertical component r :
y  d sin   140 m  sin 5.5  14 m
2
Picture the Problem: The arrow falls below the target center as it flies
from the bow to the target.
Strategy: Treat the vertical and horizontal motions separately. First
find the time required for the arrow to drop straight down 52 cm from
rest. Then use that time together with the horizontal target distance to
find the horizontal speed of the arrow. That must also equal the initial
speed because the arrow was launched horizontally.
16.
Solution: 1. Find the time to drop 52 cm:
t 
2. Find the speed of the arrow from the
horizontal distance and time elapsed:
v0 x 
0.52 m
2  0.52 m 
2y

 0.326 s
g
9.81 m/s2
x
15 m

 46 m/s
t 0.326 s
Picture the Problem: The trajectory of the clam is indicated in the figure at right.
Strategy: Use equation 4-7 to analyze the motion of the clam as it is launched horizontally and falls to the rocks below. Let
upward be the positive y direction.
Solution: 1. (a) The horizontal velocity remains constant at 2.70 m/s.
2. (b) The vertical speed increases due to the
acceleration of gravity:
v y   gt    9.81 m/s 2   2.10 s 
 –20.6 m s
19.
Picture the Problem: The pumpkin’s trajectory is depicted in the figure at right.
Strategy: Because the pumpkin’s initial velocity is horizontal, we can use equation 4-8 to find the required initial speed of the
pumpkin.
Solution: Solve equation
4-8 for v0 :
g x2

2h  y
v0 
 9.81 m/s   3.5 m 
2
2
2  9.0  0 m 
 2.6 m/s
24.
Picture the Problem: The basketball’s trajectory is depicted at right.
Strategy: Use equation 4-7 to find the x and y positions of the basketball as a
function of time. Use the right triangle formed by the floor and the
basketball’s release and landing points to write a ratio that allows us to
calculate the time of flight and therefore the initial height.
Solution: 1. Find the y position
as a function of time:
32.
y  h  12 gt 2  0
h  12 gt 2
2. Find the x position as a function of time:
x  v0 t
3. Use the tangent function for the right triangle:
tan  
4. Now solve for the flight time t:
t
5. Find the initial height:
h  12 gt 2  12  9.81 m/s2   0.495 s   1.20 m
h 12 gt 2
gt


x
v0 t
2v0
2v0 tan  2  4.20 m/s  tan 30.0

 0.495 s
g
9.81 m/s 2
2
Picture the Problem: The ball travels along a parabolic arc, maintaining its horizontal velocity but changing its vertical speed
due to the constant downward acceleration of gravity.
Strategy: The given angle of the kick allows us to calculate the vertical component of the initial velocity by using the sine
function. The time it takes the acceleration of gravity to slow down the vertical speed, bring it to zero, and speed it up again to
its initial value is the same as the time the ball is in the air.
Solution: 1. Find the y component of the initial velocity:
v0 y  v0 sin    9.85 m/s  sin 35.0  5.65 m/s
2. Let v y  v0 y and use equation 4-6 to find the time of flight:
t
v y  v0 y
g

 5.65   5.65 m/s 
9.81 m/s 2
 1.15 s
39.
Picture the Problem: The football travels along a parabolic arc, landing at the same level from which it was launched.
Strategy: The time of flight of a projectile that lands at the same level it is launched is determined by the time it takes the
acceleration of gravity to slow down the vertical component of the initial velocity to zero and then speed it up again back to its
original value. Thus upon landing the speed of the ball is v y  v0 y  v0 sin  . Use these facts to determine the time of flight and
then solve for v0 .
43.
v y  v0 y
Solution: 1. Find the time of flight:
t
2. Solve for v0 :
v0 
g

v0 sin   v0 sin  2v0 sin 

g
g
9.81 m/s2   4.50 s   24.8 m/s
gt

2sin 
2sin 63.0
Picture the Problem: The trajectory of the girl is depicted at right.
Strategy: Use equation 4-10 and the given time of flight, initial speed, and launch
angle to determine the initial height of the girl at the release point.
Solution: Use equation 4-10 to find
the initial height of the girl at the release
point. If we let the release height
correspond to y = 0, then the landing
height is:
68.
y   v0 sin   t  12 gt 2
  2.25 m/s  sin 35.0  0.616 s   12  9.81 m/s 2   0.616 s 
y  1.07 m
In other words, she was 1.07 m above the water when she let go of
the rope.
2
Picture the Problem: The seed pod is launched at 60.5° above horizontal from a location 0.455 m above the ground and follows
a parabolic arc until it hits the ground.
Strategy: Use equation 4-10 to find the vertical component of the final velocity v y . Then use the known v0 y and v y together
with the acceleration of gravity to determine the time of flight. Use the time of flight and the known horizontal speed to
determine the horizontal distance it covers during the flight.
Solution: 1. (a) First calculate v0 y :
v0 y  v0 sin    2.62 m/s  sin 60.5  2.28 m/s
2. Find v y from equation 4-6:
vy2  v02y  2 g y
vy  
 2.28 m/s 
v y  v0 y
 2  9.81 m/s2   0.455 m   3.76 m/s
3.76  2.28 m/s
 0.616 s
 9.81 m/s 2
3. Use v0 y and v y to find the time of flight:
t
4. (b) Use equation 4-10 to find x:
x   v0 cos   t   2.62 m/s  cos 60.5  0.616 s   0.795 m
g

2