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. `10 ms^(-1)`
B. `14 ms^(-1)`
C. `20 ms^(-1)`
D. `28 ms^(-1)`

Explanation

Let ball rebound with speed `v`. So
`v = sqrt(2gh) = sqrt(2 xx 10 xx 20) = 20 m//s`
Energy just after rebound
`E = (1)/(2) xx m xx v^(2) = 200 m`
`50%` energy loses in collision means just before collision energy is `400 m`
By using energy conservation.
`(1)/(2) mv_(0)^(2) + mgh = 400m`
`implies (1)/(2) mv_(0)^(2) + m xx 10 xx 20 = 400 implies v_(0) = 20 m//s`