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

From energy conservation –
`[1/2 mv^2 + "mgh" ] xx50/100 = mgh`
`implies 1/sqrt2v^2 = gh implies v = sqrt(2gh) = sqrt(2xx10 xx20) = 20 m//s`