A particle moving in a circle of radius R with a uniform speed takes a time T to complete one revolution.
If this particle were projected with the same speed at an angle '$\theta$' to the horizontal, the maximum height attained by it equals $4$R. The angle of projection, $\theta$, is then given by :
Answer: (A) $\theta = \sin^{-1}\left(\dfrac{2gT^2}{\pi^2R}\right)^{1/2}$
Concept. The circular motion gives the speed $v = \dfrac{2\pi R}{T}$; then use $H = \dfrac{v^2\sin^2\theta}{2g}$.
Step 1. Set $H = 4R$.
$$\sin^2\theta = \frac{8gR}{v^2} = \frac{8gRT^2}{4\pi^2R^2} = \frac{2gT^2}{\pi^2R}$$
Step 2. Angle.
$$\theta = \sin^{-1}\left(\frac{2gT^2}{\pi^2R}\right)^{1/2}$$
Answer. $\theta = \sin^{-1}\left(\dfrac{2gT^2}{\pi^2R}\right)^{1/2}$ (Option A).
Solution by Sreeraj P, M.Sc Physics