Question

A 0.60kg basketball is dunked vertically downward so that it leaves the [layer's hand at 8m/s. Assume it is released when the bottom of the ball is at the rim (3.05 meters off the ground).

a). find acceleration of the ball once it leaves the player's hand
b). how ling will it take the ball to reach the ground.
c). what will be the velocity of the ball when it hit the ground.

Answer 1

b. h = Vo*t + 0.5g*t^2 = 3.05

8t + 4.9t^2 - 3.05 = 0
Use Quadratic Formula and get:
t = 0.319 s.

c. V^2 = Vo^2 + 2g*h
V^2 = 8^2 + 19.6*3.05 = 123.78
V = 11.13 m/s.