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

Time period of circular motion is given by, 

$\dfrac{2\piR}{v} = T ⇒ v = \dfrac{2\piR}{T}$

Now, maximum height $H = \dfrac{{v}^{2}\sin^{2}\theta}{2g}$

$⇒ 4R = \dfrac{4{\pi}^{2}{R}^{2}}{{T}^{2}}\dfrac{\sin^{2}\theta}{2g}$

$⇒ sin\theta = \sqrt{\dfrac{2g{T}^{2}}{{\pi}^{2}R}}$

$⇒ \theta = \sin^{- 1}\left(\dfrac{2g{T}^{2}}{{\pi}^{2}R}\right)\dfrac{1}{2}$