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 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?
1 answer