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

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$θ = \sin^{- 1} {(\frac{2 {gT}^{2}}{π^{2} R})}^{1 / 2}$

$θ = \cos^{- 1} {(\frac{{gT}^{2}}{π^{2} R})}^{1 / 2}$

$θ = \cos^{- 1} {(\frac{π^{2} R}{{gT}^{2}})}^{1 / 2}$

$θ = \sin^{- 1} {(\frac{π^{2} R}{{gT}^{2}})}^{1 / 2}$

answer is A.

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Detailed Solution

Velocity of the particle $V = \frac{2 πR}{T}$
When projected at angle $θ$
H = maximum height $= \frac{v^{2} \sin^{2} θ}{2 g} = 4 R$
$\begin{array}{l} = \frac{4 π^{2} R^{2}}{T^{2}} \frac{\sin^{2} θ}{2 g} = 4 R \\ \sin^{2} θ = \frac{2 {gRT}^{2}}{π^{2} R^{2}} \\ \sin^{2} θ = \frac{2 {gT}^{2}}{π^{2} R} \\ \sin θ = {(\frac{2 {gT}^{2}}{π^{2} R})}^{1 / 2} \\ θ = \sin^{- 1} {(\frac{2 {gT}^{2}}{π^{2} R})}^{1 / 2} \end{array}$