Question

At what angle should a ball be projected up an inclined plane with a velocity $v_0$, so that it may hit the incline normally. The angle of the inclined plane with the horizontal is $\alpha$.

Answer 1

As the ball has to hit the inclined plane normally, in that position the $x$-component of velocity will be zero and the 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,

$v_x=u_x+a_{x}t$

Here, $v_x=0,u_x=v_{0}cos\theta ,a_x=-gsin\alpha$

$0=v_{0}cos\theta -\left(gsin\alpha \right)T\Rightarrow T=\frac{v_{0}cos\theta }{gsin\alpha }$ ....(i)

Also the displacement of the particle in y-direction will be zero. Using

$y=u_{y}t+\frac{1}{2}{a}_y{t}^2\Rightarrow 0=v_{0}sin\theta T-\frac{1}{2}gcos{T}^2$

This gives $T=\frac{2v_{0}sin\theta }{gcos\alpha }$ ...(ii)

From (i) and (ii), we have

$\frac{v_{0}cos\theta }{gsin\alpha }=\frac{2v_{0}sin\theta }{gcos\alpha }$

$\Rightarrow \frac{cos\theta }{sin\alpha }=\frac{2sin\theta }{cos\alpha }$

$tan\theta =\left[\frac{1}{2}cot\alpha \right]$

$\theta =ta{n}^{-1}\left(\frac{1}{2}cot\alpha \right)$

Which is the required angle of projection.