Question
A baseball pitcher throws a ball vertically upward and catches it at the same
level 4.2 s later.
(a) With what velocity did the pitcher throw the ball?
(b) How high does the ball rise?
level 4.2 s later.
(a) With what velocity did the pitcher throw the ball?
(b) How high does the ball rise?
Answers
GPT 3.5
To solve this problem, we can use the equations of motion for vertical motion:
(a) The initial velocity can be found using the equation: v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
In this case, the final velocity is 0 m/s (since the ball stops when it reaches its maximum height), the acceleration is -9.8 m/s^2 (due to gravity acting in the opposite direction), and the time is 4.2 seconds.
Therefore, we can rearrange the equation to solve for the initial velocity:
0 = u + (-9.8)(4.2)
0 = u - 41.16
u = 41.16 m/s
Therefore, the pitcher threw the ball upwards with an initial velocity of 41.16 m/s.
(b) The maximum height can be found using the equation: s = ut + 0.5at^2, where s is the displacement (change in position), u is the initial velocity, a is the acceleration, and t is the time.
In this case, the displacement is 0 m (since the ball returns to the same level), the initial velocity is 41.16 m/s, the acceleration is -9.8 m/s^2, and the time is 4.2 seconds.
Therefore, we can rearrange the equation to solve for the maximum height:
0 = (41.16)(4.2) + 0.5(-9.8)(4.2)^2
0 = 172.872 + 0.5(-9.8)(17.64)
0 = 172.872 - 85.584
h = 87.288 m
Therefore, the ball rises to a height of 87.288 meters.
(a) The initial velocity can be found using the equation: v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
In this case, the final velocity is 0 m/s (since the ball stops when it reaches its maximum height), the acceleration is -9.8 m/s^2 (due to gravity acting in the opposite direction), and the time is 4.2 seconds.
Therefore, we can rearrange the equation to solve for the initial velocity:
0 = u + (-9.8)(4.2)
0 = u - 41.16
u = 41.16 m/s
Therefore, the pitcher threw the ball upwards with an initial velocity of 41.16 m/s.
(b) The maximum height can be found using the equation: s = ut + 0.5at^2, where s is the displacement (change in position), u is the initial velocity, a is the acceleration, and t is the time.
In this case, the displacement is 0 m (since the ball returns to the same level), the initial velocity is 41.16 m/s, the acceleration is -9.8 m/s^2, and the time is 4.2 seconds.
Therefore, we can rearrange the equation to solve for the maximum height:
0 = (41.16)(4.2) + 0.5(-9.8)(4.2)^2
0 = 172.872 + 0.5(-9.8)(17.64)
0 = 172.872 - 85.584
h = 87.288 m
Therefore, the ball rises to a height of 87.288 meters.