Question

A ball is thrown vertically downwards from a height of 20 m with a initial velocity v0. It colloids with the ground, loses 50 % of its energy in collision and rebounds to the same height. The initial velocity v0 is .....??? Please explain it briefly.

Answer 1

`v_0 = 19.82 m/s`

Explanation:

v = v0 + g t
h = v0 t + g t²/2 = 20
(Everything measured downwards)

`E_k` = (1/2) m v²
=> `E_p` = m g h = 196.2 m = (1/4) m v²
=> v ² = 784.8
=> v = 28.014 m/s

28.014 = v0 + 9.81 t
20 = v0 t + 4.905 t²

`=>` 20 = (28.014 - 9.81 t) t + 4.905 t²
= 28.014 t - 4.905 t²
`=>` 4.905 t² - 28.014 t + 20 = 0
This is a quadratic equation in t
t = (28.014 `\pm` 19.809)/9.81
The solution with the - sign yields
t = 0.836 s `=>` v0 = 19.82 m/s