A body is projected up with a speed `v_0` along the line of greatest slope of an inclined plane of angle of inclination `beta`. If the body collides elastically perpendicular to the inclined plane, find the time after which the body passes through its point of projection.

Explanation

On striking the plane at `B`, the velocity along the x - axis should be zero. So using `v_x = u_x + a_x t`, we get
`0 =m v_0 cos prop - g sin beta T`, where `T` is the time of flight
`T = (v_0 cos prop)/(g sin beta)`…(i)
Also `S_y = 0` from A to B. So using `S_y = u_y t + (1)/(2) a_y t^2`, we get
`0 = (v_0 sin prop) T - (1)/(2) (g cos beta) T^2 rArr T = (2 v_0 sin prop)/( g cos beta)` ...(ii)
From (i) and (ii), `(v_0 cos prop)/(g cos beta) = (2 v_0 sin prop)/(g cos beta) rArr tan prop = (1)/(2) cot beta`
`rArr cos prop = (2)/(sqrt(4 + cot^2 beta)) = (2 sin beta)/(sqrt(2 + 3 sin^2 beta))`
Put the value of `cos prop` in (i) to get `T = (2 v_0)/(g sqrt(1 + 3 sin^2 beta))`
Since the body collides elastically, it will rebound perpendicular to the inclined plane with the same speed `v`. In consequence, it retraces its path elapsing equal time `T` for backward journey to the point of projection. Hence, the required time of motion is
`T' = 2 T = (4 v_0)/(g sqrt(1 + 3 sin^2 beta))`.
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