Q. 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 4R. The angle of projection, $\theta$, is then given by:
A) $\theta = \cos^{-1}\left(\frac{\pi^2R}{gT^2}\right)^{1/2}$
B) $\theta = \sin^{-1}\left(\frac{\pi^2R}{gT^2}\right)^{1/2}$
C) $\theta = \sin^{-1}\left(\frac{2gT^2}{\pi^2R}\right)^{1/2}$
D) $\theta = \cos^{-1}\left(\frac{gT^2}{\pi^2R}\right)^{1/2}$
Explanation
[Solution] $v = 2\pi R/T$; $H = \frac{v^2\sin^2\theta}{2g} = 4R$ gives $\sin^2\theta = \frac{8gR}{v^2} = \frac{2gT^2}{\pi^2R}$. Answer (3).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.