A particle moving in a circle of radius $R$ with a uniform speed takes a time $T$ to complete one revolution.

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

Accepted Answer

To complete a circular path of radius $R$, time period is $T$. So speed of particle $U = \frac{2\pi R}{T}$. Now the particle is projected with same speed at angle $	heta$ to horizontal. Maximum Height $H = \frac{U^2 \sin^2 	heta}{2g}$. Given $H = 4R$, so $\frac{U^2 \sin^2 	heta}{2g} = 4R$. Substituting $U = \frac{2\pi R}{T}$, we get $\frac{1}{2g} \left( \frac{4\pi^2 R^2}{T^2} ight) \sin^2 	heta = 4R$. Simplifying, $\sin^2 	heta = \frac{8gR T^2}{4\pi^2 R^2} = \frac{2gT^2}{\pi^2 R}$. Thus, $	heta = \sin^{-1} \left( \frac{2gT^2}{\pi^2 R} ight)^{1/2}$.

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