Question

A ball is dropped from rest onto a fixed horizontal surface and rebounds to a height which is 64% of its original height. (a) What is the coefficient of restitution? (b) With what vertical velocity must the ball strike the surface to rebound to a height of 25 ft?

Answer 1

(a) The coefficient of restitution is 0.8.

(b) The ball must strike the surface with a 50.14 ft/s vertical velocity, approximately.

Step-by-step explanation:

The ratio between the final and the initial height is given by

`h_f = 0.64 h_i`

Before the impact, we have

`v_f^2 = v_0^2 + 2gh_i\\v_f = √(2gh_i).`

Analogously after the impact

`v_f^2 = v_0^2 - 2gh_f\\v_0 = √(2gh_f).`

From the previous, the restitution coefficient can be calculated from the ratio between the ball's velocity, after and before the impact, i.e.

`(v_0)/(v_f) = (√(2gh_f))/(√(2gh_i)) = \sqrt{(h_f)/(h_i)} = \sqrt{(0.64 h_i)/(h_i)} = √(0.64) = \mathbf{0.8}.`

If the ball is meant to rebound to a 25 ft height, we plug in this value in one of the previous equations as follows

`v_f = √(2gh_i) = \sqrt{2g(h_f)/(0.64)} = \sqrt{2* 32.18(ft)/(s^2) *(25 ft)/(0.64)} \approx \mathbf{50.14 ft/s}.`