Question

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 ' $	heta$ ' to the horizontal, the maximum height attained by it equals 4R. The angle of projection, $	heta$, is then given by:

Accepted Answer

$	heta = \sin^{- 1}\left( \frac{{2gT}^{2}}{\pi^{2}R} ight)^{\frac{1}{2}}$

Answer

$	heta = \cos^{- 1}\left( \frac{{gT}^{2}}{\pi^{2}R} ight)^{\frac{1}{2}}$

Answer

$	heta = \cos^{- 1}\left( \frac{\pi^{2}R}{{gT}^{2}} ight)^{\frac{1}{2}}$

Answer

$	heta = \sin^{- 1}\left( \frac{\pi^{2}R}{{gT}^{2}} ight)^{\frac{1}{2}}$

Explanation

Practice

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