A particle moving in a circle of radius R with a uniform speed takes 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:

(1) $\theta = {\cos ^{ - 1}}\sqrt {\frac{{g{T^2}}}{{{\pi ^2}R}}} $

(2) $\theta = {\cos ^{ - 1}}\sqrt {\frac{{{\pi ^2}R}}{{g{T^2}}}} $

(3) $\theta = {\sin ^{ - 1}}\sqrt {\frac{{{\pi ^2}R}}{{g{T^2}}}} $

(4) $\theta = {\sin ^{ - 1}}\sqrt {\frac{{2g{T^2}}}{{{\pi ^2}R}}} $

*Solution*

For circular motion we have, $v=\frac {2\pi R}{T}$

For projectile motion we have, max. height = $H=\frac {v^2 sin^2 \theta }{2g}=4R$

$ \Rightarrow {\left( {\frac{{2\pi R}}{T}} \right)^2}\frac{{{{\sin }^2}\theta }}{{2g}} = 4R$

$\therefore \sin \theta = \sqrt {\frac{{2g{T^2}}}{{{\pi ^2}R}}} $ Or $\theta = {\sin ^{ - 1}}\sqrt {\frac{{2g{T^2}}}{{{\pi ^2}R}}} $

Answer: (4)