At what angle should a ball be projected up an inclined plane with a velocity so that it may hit the incline normally. The angle of the inclined plane with the horizontal is α.
θ=tan−1(12cot α)
As the ballhas to hit the inclined plane normally, so in that position the x-component of velocity will be zero and velocity will have y-component only.
The ball will hit the incline normally if its parallel component of velocity reduces to zero during the time of flight.
By analyzing this motion along incline, i.e., x-direction vx=ux+axt
Here vx=0,ux=v0cos θ,ax=−gsin α
0=vocos θ−(gsin α)T⇒T=vocos θgsin α ........(i)
Also the displacement of the particle in y-direction will be zero. Using
y=uyt+12ayt2⇒0=vosin θ.T−12gcos α.T2
This gives T = 2vosin θgcos α ...............(ii)
From (i) and (ii), we have vocos θgsin α=2vosin θgcos α⇒cos θsin α=2sin θcos α⇒2tan θtan α=1⇒tan θ=[12cos α]
⇒=tan−1(12cot α) which is the required angle of projection.