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

• A. $\theta=\cos ^{-1}\left(\frac{ gT ^{2}}{\pi^{2} R }\right)^{1 / 2}$
• B. $\theta=\cos ^{-1}\left(\frac{\pi^{2} R}{g T^{2}}\right)^{1 / 2}$
• C. $\theta=\sin ^{-1}\left(\frac{\pi^{2} R}{ g T ^{2}}\right)^{1 / 2}$
• D. $\theta=\sin ^{-1}\left(\frac{2 gT ^{2}}{\pi^{2} R }\right)^{1 / 2}$

Answer 1

Answer: (D)

$\theta=\sin ^{-1}\left(\frac{2 gT ^{2}}{\pi^{2} R }\right)^{1 / 2}$

Answer 2

Velocity of the particle $v=\frac{2 \pi R}{T}$ When projected at angle $\theta$ $H=$ maximum height $=\frac{v^{2} \sin ^{2} \theta}{2 g}=4 R$ $=\frac{4 \pi^{2} R^{2}}{T^{2}} \frac{\sin ^{2} \theta}{2 g}=4 R$ $\sin ^{2} \theta=\frac{2 g R T^{2}}{\pi^{2} R^{2}}$ $\sin ^{2} \theta=\frac{2 g T^{2}}{\pi^{2} R}$ $\sin \theta=\left(\frac{2 g T^{2}}{\pi^{2} R}\right)^{1 / 2}$ $\theta=\sin ^{-1}\left(\frac{2 g T^{2}}{\pi^{2} R}\right)^{1 / 2}$