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?

A ball falls on an inclined plane of inclinatioin theta from a height

1.	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» Solution :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 ata distance.