Atennis player tosses a tennis ball straight up and then catches it after 1.62 s at the same height as the point of release. (a) what is the acceleration of the ball while it is in flight? (b) what is the velocity of the ball when it reaches its maximum height? (c) find the initial velocity of the ball. (d) find the maximum height it reaches.

A. When the ball is in flight, the acceleration would be = - 9.8 m/sec^2. The reasoning being that it is going against gravity.

b. When the ball reaches its maximum height, the velocity of the ball will be zero.

c. Let us assume the initial velocity of the ball = v1
We already know that final velocity of the ball (y) = 0 = x
Then
v1 = (1/2) * (9.8) * (1.62)
     = 4.9 * 1.62
     = 7.9 m/s^2
d. The velocity of the ball at maximum height is zero.
Let us assume the maximum height reached by the ball = h
Then
v^2 = - 2gh
(7.9)^2 = -2 * (-9.8) * h
62.41 = 19.6 h
h = 62.41/19.6
   = 3.2 m

Are there any statements?

X=y+4 is the answe