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 m//s`
B. `14 m//s`
C. `20 m//s`
D. `28 m//s`

Explanation

Here, `h=20m, g=10 m//s^(2), u=upsilon_(0)`
Initial energy `=mgh+(1)/(2)m upsilon_(0)^(2)`
final energy `=mgh(` As ball rebounds to the same height `)`
According to question
`(1)/(2)((1)/(2)m upsilon_(0)^(2)+mgh)=mgh`
`(1)/(2)m ((1)/(2)upsilon_(0)^(2)+gh)=mgh` or ` (1)/(4)upsilon_(0)^(2)+(1)/(2)gh=gh`
or `(1)/(4)upsilon_(0)^(2)=(1)/(2)gh ` or ` upsilon_(0)^(2)=2gh`
`upsilon_(0)=sqrt(2gh)=sqrt(2xx10xx20)=20m//s`