Q 11-03-058JEE MainJEE Main 2024 (1 Feb, Shift 1)Medium
A particle moving in a circle of radius $R$ with uniform speed takes time $T$ to complete one revolution. If this particle is projected with the same speed at an angle $\theta$ to the horizontal, the maximum height attained by it is equal to $4R$. The angle of projection $\theta$ is then given by
Answer: (A) $\sin^{-1}\left(\dfrac{2gT^2}{\pi^2 R}\right)^{1/2}$
Speed in the circle: $v = \dfrac{2\pi R}{T}$.
Maximum height of the projectile:
$$H = \frac{v^2\sin^2\theta}{2g} = 4R \Rightarrow \sin^2\theta = \frac{8gR}{v^2} = \frac{8gR\,T^2}{4\pi^2R^2} = \frac{2gT^2}{\pi^2R}$$
$$\theta = \sin^{-1}\left(\frac{2gT^2}{\pi^2R}\right)^{1/2}$$
Solution by Sreeraj P, M.Sc Physics