A ball is thrown vertically downwards from a height of 20 m with an initial velocity `v_(0)` It collides with the ground , loses 50 percent of its energy in collision and rebounds to the same height .the initial velocity `v_(0)` is : (Take `g = 10 ms^(-2)`)

Options

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

Answer

C

Explanation

Let ball rebounds with speed v so
` :. "In" v^(2) -v_(0)^(2) =2gh ,v_(0) = 0 `
` :. V= sqrt(2gh) =sqrt(2xx10xx20) = 20 m//s `
Energy just after rebound
`E =1/2 mv^(2) =1/2 m xx m xx (20)^(2)`
` = 1/2 m xx400`
= 200 m Joule
50 % energy loses in collision means just before collision energy is 400 m
By using energy conservation
`1/2 mv_(0)^(2) +mgh = 400 m `
` :. v_(0)^(2) = 800 - 2gh`
` :. v_(0)^(2) = 800 - 400`
` :. v_(0)^(2) = 400 `
` :. v_(0) = 20 ms^(-1)`