Question

A ball is dropped from a height of 1.20 m and hits the floor. The ball compresses and then reforms to spring upwards from the floor to a height of 0.86 m. The time from the ball's first contact to when it left contact with the floor was 0.091 s. Calculate the ball's acceleration while it is in contact with the floor.

Answer 1

The ball's acceleration while it is contact with the floor is 98.4m/s²

Explanation:

V1 = -√(2gh1) = -4.85 m/s

V2 = √(2gh2) = 4.1056 m/s

dV = V2 - V1 = +8.9556 m/s

a = dV/dt = 8.9556/0.091 = 98.4 m/s²

Therefore, the ball's acceleration while it is contact with the floor is 98.4m/s²