A ball is projected vertically down with an initial velocity from a height of `20 m` onto a horizontal floor. During the impact it loses `50%` of its energy and rebounds to the same height. The initial velocity of its projection is

Options

A. `20 ms^(-1)`
B. `15 ms^(-1)`
C. `10 ms^(-1)`
D. `5 ms^(-1)`

Answer

A

Explanation

Let ball is projected vertically downward with velocity v from height h
Total energy at point `A= (1)/(2) mv^(2)+ mgh`
During collision loss of energy is 50% and the ball rises up to same height. It means it possess only potential energy at same level
`50% ((1)/(2) mv^(2) + mgh)= mgh`
`(1)/(2) ((1)/(2) mv^(2) + mgh)= mgh`
`v= sqrt(2gh)= sqrt(2 xx 10 xx 20)`
`therefore v= 20 m//s`