A ball released from a height ho above a horizontal surface rebounds to a height `h_(1)`, after one bounce. The graph that relates `h_(0)` to `h_(1)` is shown Fig. If the ball (of the mass `m`) was dropped from an initial height `h` and made three bounces, the kinetic energy of the ball immediately after the third impact with the surface was

Options

A. `(0.8)^(3)mgh`
B. `(0.8)^(2)mgh`
C. `0.8mg(h//3)`
D. `[1-(0.8)^(3)]mgh`

Answer

A

Explanation

`h_(1)/h_(0)=e^(2)implies 80/100=e^(2)impliese=sqrt0.8`
velocity before the first impact `u=sqrt(2gh)`
velocity after the third impact `v_(3)=e^(3)sqrt(2gh)`
`KE=1/2mv_(3)^(2)=e^(6)mgh=(0.8)^(3)mgh`