At what angle should a ball be projected up an inclined plane with a velocity `v_0` so that it may hir the incline normally. The angle of the inclined plane with the horozontal is `prop`.
.

Explanation

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., s - direction,
`v_x = u_x + a_x t`
Here `v_x = 0, u_x = v_0 cos theta, a_x = -g sin prop`
`0 = v_0 cos theta - (g sin prop) T rArr T = (v_0 cos theta)/(g sin prop)`...(i)
Also the displacement of the particle in y - direction will be zero. Using
`y = u_y t + (1)/(2) a_y t^2 rArr 0 = v_0 sin theta T - (1)/(2) g cos T^2`
This gives `T = (2 v_0 sin theta)/(g cos prop)`
From (i) and (ii), we have
`(v_0 cos theta)/(g sin prop) = (2 v_0 sin theta)/(g cos prop)`
`rArr (cos theta)/(sin prop) = (2 sin theta)/(cos prop)`
`2 tan theta tan prop = 1`
`tan theta = [(1)/(2) cot prop]`
`theta = tan^-1 ((1)/(2) cot prop)`
Which is the required angle of projection.