A ball falls on an inclined plane of inclinatioin `theta` from a height h above the pontof impact and makes a perfectly elastic collision. Where will it hit the plane again?

Answer

A, C

Explanation

The ball stikes the inclined plane at origiln with velocity `v_0=sqrt(2gh)`. As the ball elastically rebounds, it recalls wilth same velocity `v_0` at the sme angle `theta from the normakl or y-axis. Let the ball strikes the incline second time at any pointnP which is at distance l from the origin along teh incline. From teh equation
`y=v_(iy)t+1/2w_yt^2`
`0=v-0costhetast-1/2gcosthetat^2`
where t is th same time of motion of ball in air which moving from orign to P.
`As t!=0 , so t=(2v_0)/g`

Now from the equation ltbr. `x+v_(0x)t+1/2w_xt^2`
l=v_0sinthetat+1/2gsinthetat^2`
so, `l=v_0sintheta((2v_0)/g)+1/2gsintheta((2v_0)/g)^2`
`=(2v_0^2sintheta)/g`
Hence the plane will hit again at a distance.