A particle moving in a circle of radius R wit | vedclass

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(D) The speed of the particle in circular motion is $V = \frac{2 \pi R}{T}$.The maximum height $H$ attained by a projectile is given by $H = \frac{V^2

Vedclass · Accepted Answer

$	heta=\sin ^{-1}\left(\frac{2 g T^{2}}{\pi^{2} R}ight)^{1 / 2}$

Vedclass · Answer

(D) The speed of the particle in circular motion is $V = \frac{2 \pi R}{T}$.The maximum height $H$ attained by a projectile is given by $H = \frac{V^2 \sin^2 	heta}{2g}$.Given $H = 4R$,we substitute $V$ and $H$ into the formula:$4R = \frac{(\frac{2 \pi R}{T})^2 \sin^2 	heta}{2g}$$4R = \frac{4 \pi^2 R^2 \sin^2 	heta}{2g T^2}$$4R = \frac{2 \pi^2 R^2 \sin^2 	heta}{g T^2}$Solving for $\sin^2 	heta$:$\sin^2 	heta = \frac{4R \cdot g T^2}{2 \pi^2 R^2} = \frac{2 g T^2}{\pi^2 R}$Therefore,$	heta = \sin^{-1} \left( \frac{2 g T^2}{\pi^2 R} ight)^{1/2}$.

$A$ 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

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