A particle moving in a circle of radius $\mathrm{R}$ 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 :

<p>To solve for the angle of projection $\theta$, we will first establish the relationship between the variables given and then derive the formula using kinematics.</p> <p>The time $T$ for one revolution at speed $v$ in a circle of radius $R$ is related to the circumference of the circle by the formula:</p> <p>$v = \frac{2\pi R}{T}$</p> <p>When the particle is projected with the same speed $v$ at an angle $\theta$ to the horizontal, its vertical component of velocity is given by $v_y=v\sin\theta$.</p> <p>The maximum height $H$ reached by the projectile can be found from the kinematic equation:</p> <p>$H = \frac{v_y^2}{2g} = \frac{(v\sin\theta)^2}{2g}$</p> <p>We are given that the maximum height attained $H$ is equal to $4R$, so:</p> <p>$4R = \frac{(v\sin\theta)^2}{2g}$</p> <p>Substitute $v$ from the first equation into the second one:</p> <p>$4R = \frac{((\frac{2\pi R}{T})\sin\theta)^2}{2g}$</p> <p>$4R = \frac{(2\pi R\sin\theta)^2}{2gT^2}$</p> <p>$4R = \frac{4\pi^2 R^2 \sin^2\theta}{2gT^2}$</p> <p>$2gT^2 = \pi^2 R \sin^2\theta$</p> <p>Now solve for $\sin\theta$:</p> <p>$\sin\theta =\left[\frac{2gT^2}{\pi^2 R}\right]^{1/2}$</p> <p>Finally, to solve for $\theta$, take the inverse sine of both sides:</p> <p>$\theta = \sin^{-1}\left(\left[\frac{2gT^2}{\pi^2 R}\right]^{1/2}\right)$</p> <p>Therefore, the correct answer is:</p> <p>Option A</p> <p>$\sin^{-1}\left[\frac{2gT^2}{\pi^2 R}\right]^{1/2}$</p>