A ball is falling freely from a certain height. When it reached `10 m` height from the ground its velocity is `v_(0)`. It collides with the horizontal ground and loses `50 %` of its energy and rises back to height of `10 m`. The value of velocity `v_(0)` is

Options

A. 7 m/s
B. 10 m/s
C. 14 m/s
D. 16 m/s

Answer

C

Explanation

The velocity of the body on reaching the ground is
`V = sqrt(u^(2) + 2gh) = sqrt(V_(0)^(2) + 196)`
Given that,
`(1)/(2) (KE) = PE`
`(1)/(2) ((1)/(2) mV^(2)) = mgh`
`V^(2) = 4 gh`
`V_(0)^(2) + 196 = 4 xx 9.8 xx 10`
`V_(0)^(2) = 392 - 196 = 196`
`V_(0) = 14 m//s`