Question

A ball falls on an inclined plane of inclination theta from a height h above the point of impact

Answer 1

so what's the question u want to ask ???

Answer 2

The ball stikes the inclined plane at origiln with velocity v0=2gh−−−√. As the ball elastically rebounds, it recalls wilth same velocity v0 at the sme angle θomthe∥a∥klory−aξs.Lettheballstrikesthe∈cl∈esecondtimeatanyp∮nPwhichisatdistancelomtheorig∈alongteh∈cl∈e.Fromtehequationy=v_(iy)t+1/2w_yt^20=v-0costhetast-1/2gcosthetat^2wheretisthsametimeofmotionofball∈airwhichmov∈gomorign→P.As t!=0 , so t=(2v_0)/g`

Now from the equation ltbr. x+v0xt+12wxt2

l=v_0sinthetat+1/2gsinthetat^2so,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.

I hope it is helpful

please mark me as brainliest