Question

you toss a racquetball directly upward and then catch it at the same height you released it 1.90 s later. Assume air resistance is negligible.What is the acceleration of the ball while it is moving upward?What is the acceleration of the ball while it is moving downward?What is the acceleration of the ball while it is at its maximum height?What is the velocity of the ball when it reaches its maximum height?What is the initial velocity of the ball?What is the maximum height that the ball reaches?

Answer 1

the acceleration is about 9.81 m/s^2 down the entire trip

velocity is zero at the top. It stops then falls.

it went up for 1.9/2 = 0.45 s
v = Vi - g t
at top v = 0
0 = Vi - 9.81 (.45)
so Vi = 9.81*.45

h = Vi (.45) - 4.9 (.45)^2
h is height above your hand