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Question

A particle moving in a circle of radius $R$ with uniform speed takes time $T$ to complete one revolution. If this particle is projected with the same speed at an angle $	heta$ to the horizontal, the maximum height attained by it is equal to $4R$. The angle of projection $	heta$ is then given by: A). $\sin^{-1}\!\left[\dfrac{2gT^2}{\pi^2 R}ight]^{\!\frac{1}{2}}$ B). $\sin^{-1}\!\left[\dfrac{\pi^2 R}{2gT^2}ight]^{\!\frac{1}{2}}$ C). $\cos^{-1}\!\left[\dfrac{2gT^2}{\pi^2 R}ight]^{\!\frac{1}{2}}$ D). $\cos^{-1}\!\left[\dfrac{\pi R}{2gT^2}ight]^{\!\frac{1}{2}}$

Edvaya · Accepted Answer

The correct answer is $\sin^{-1}\!\left[\dfrac{2gT^2}{\pi^2 R}ight]^{\!\frac{1}{2}}$. Speed of particle in circular motion: $v = \dfrac{2\pi R}{T}$
Maximum height of projectile: $H = \dfrac{v^2\sin^2	heta}{2g} = 4R$
Substituting $v$:
$\dfrac{4\pi^2 R^2}{T^2} \cdot \dfrac{\sin^2	heta}{2g} = 4R$
$\sin^2	heta = \dfrac{8gRT^2}{4\pi^2 R^2} \cdot \dfrac{1}{1} = \dfrac{2gT^2}{\pi^2 R}$
$\sin	heta = \left[\dfrac{2gT^2}{\pi^2 R}ight]^{\!\frac{1}{2}}$
$	heta = \sin^{-1}\!\left[\dfrac{2gT^2}{\pi^2 R}ight]^{\!\frac{1}{2}}$

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