Determine the horizontal velocity `v_0` with which a stone must be projected horizontally from a point P, so that it may hit the inclined plane perpendicularly. The inclination of the plane with the horizontal is `theta` and point P is at a height h above the foot of the incline, as shown in the figure.

Answer

A, B, C

Explanation

`u_x = v_0 cos theta, u_y = v_0sin theta, a_x = -g sin theta`,
`a_y = g cos theta`
At Q, `v_x = 0`
`:. u_x + a_xt = 0`

or `t = (v_0cos theta)/(g sin theta)` ……..(i)
`s_y = h cos theta`
`:. u_yt + 1/2 a_yt^2 = hcos theta`
`:. (v_0 sin theta)((v_0cos theta)/(g sin theta))`
`+ 1/2 (g cos theta)((v_0 cos theta)/(g sin theta)) = h cos theta`
Solving this equation we get,
`v_0 = sqrt(2gh)/(2+cot^2theta)`.