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

`(1)`

Explanation

`1/2 mv_(1)^(2) = 50/100(1/2 mv^(2))`
`v_(1)^(2) = (v^(2))/2, v `is velocity with which ball strikes ground , `v_(1)` is velocity after rebound.
When ball rises to maximum height
`0 - v_(1)^(2) = -2gh`
`(v^(2))/2 = 2gh Rightarrow v^(2) = 4gh`
When ball projected down,
`v^(2) - u^(2) = 2gh`
`4gh - u^(2) = 2gh Rightarrow u^(2) = 2gh`
`u = sqrt(2gh) = 20`