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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 $4 \mathrm{R}$. The angle of projection, $\theta$, is then given by
$\theta=\cos ^{-1}\left(\frac{g T^{2}}{\pi^{2} R}\right)^{1 / 2}$
$\theta=\cos ^{-1}\left(\frac{\pi^{2} R}{g T^{2}}\right)^{1 / 2}$
$\theta=\sin ^{-1}\left(\frac{\pi^{2} R}{g T^{2}}\right)^{1 / 2}$
$\theta=\sin ^{-1}\left(\frac{2 \mathrm{~g} T^{2}}{\pi^{2} R}\right)^{1 / 2}$
Solution
$T=\frac{2 \pi R}{V}$
$V=\frac{2 \pi R}{T}$
$H=\frac{u^{2} \sin ^{2} \theta}{2 g}$
$4 R=\frac{4 \pi^{2} R^{2} \sin ^{2} \theta}{T^{2} 2 g}$
$\sin ^{2} \theta=\frac{8 R T^{2} g}{4 \pi^{2} R^{2}}$
$\sin \theta=\sqrt{\frac{2 T^{2} g}{\pi^{2} R}}$
$\theta=\sin ^{-1}\left(\frac{2 T^{2} g}{\pi^{2} R}\right)^{1 / 2}$
