A ball which is at rest is dropped from a height h metre. As it bounces off the floor its speed is 80% of what it was just before touching the ground. The ball will then rise to nearly a height:

Options

A. `0.94 h`
B. `0.80 h`
C. `0.75 h`
D. `0.64 h`

Answer

D

Explanation

`v = sqrt (2 gh) " "...(1)`
After rebounce, `v ^(2) = u ^(2) - 2 gh`
`implies u ^(2) = v ^(2) + 2 gh`
`therefore u ^(2) = 2 gh. " "...(2)`
`therefore (v ^(2))/( u ^(2)) = ( 2 gh)/( 2 gh.) `
`implies h = h xx ( u ^(2))/(v ^(2)) = h xx ((80)/( 100)) ^(2)`
`= 0.64 h`