A particle moving in a circle of radius \(R\) with uniform speed takes time \(\mathrm{T}\) to complete one revolution. If this particle is projected with the same speed at an angle \(\theta\) to the horizontal, the maximum height attained by it is equal to \(4 R\). The angle of projection \(\theta\) is then given by _______.

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Answer & Solution

Correct Answer

(A) \(\sin ^{-1}\left[\frac{2 g T^2}{\pi^2 R}\right]^{\frac{1}{2}}\)

Step-by-step Solution

Detailed explanation

\(\frac{2 \pi R}{T}=V\)

\(\text { Maximum height } H=\frac{v^2 \sin ^2 \theta}{2 g}\)

\(4 R=\frac{4 \pi^2 R^2}{T^2 2 g} \sin ^2 \theta\)

\(\sin \theta=\sqrt{\frac{2 g T^2}{\pi^2 R}}\)

\(\theta=\sin ^{-1}\left(\frac{2 g T^2}{\pi^2 R}\right)^{\frac{1}{2}}\)

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