Question 40: 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 4R. The angle of projection, \theta, is then given by :
(1) \quad \theta=\cos ^{-1}\left(\frac{g T^{2}}{\pi^{2} R}\right)^{1 / 2}
(2) \quad \theta=\cos ^{-1}\left(\frac{\pi^{2} \mathrm{R}}{\mathrm{gT}^{2}}\right)^{1 / 2}
(3) \quad \theta=\sin ^{-1}\left(\frac{\pi^{2} \mathrm{R}}{\mathrm{gT}^{2}}\right)^{1 / 2}
(4) \quad \theta=\sin ^{-1}\left(\frac{2 \mathrm{gT}^{2}}{\pi^{2} \mathrm{R}}\right)^{1 / 2}
Answer: Option (4)
Explanation:
The speed of the particle moving uniformly in a circle of radius R and time period T is
v=\frac{2\pi R}{T}When the particle is projected with the same speed at an angle \theta,
the maximum height reached in projectile motion is given by
H=\frac{v^{2}\sin^{2}\theta}{2g}According to the question, the maximum height is 4R. Hence,
\frac{v^{2}\sin^{2}\theta}{2g}=4RSubstituting v=\frac{2\pi R}{T},
\frac{\left(\frac{2\pi R}{T}\right)^{2}\sin^{2}\theta}{2g}=4R \frac{4\pi^{2}R^{2}\sin^{2}\theta}{2gT^{2}}=4RSimplifying,
\frac{2\pi^{2}R\sin^{2}\theta}{gT^{2}}=4 \sin^{2}\theta=\frac{2gT^{2}}{\pi^{2}R}Taking square root on both sides,
\sin\theta=\left(\frac{2gT^{2}}{\pi^{2}R}\right)^{1/2}Therefore, the angle of projection is
\theta=\sin^{-1}\left(\frac{2gT^{2}}{\pi^{2}R}\right)^{1/2}Hence, the correct answer is option (4).