1. Speed in uniform circular motion

If a particle moves in a circle of radius $R$ and completes one revolution in time $T$, then its speed is

$v = \frac{2\pi R}{T}.$

2. Projectile motion with the same speed

The particle is now projected with speed $v$ at an angle $\theta$ to the horizontal.

Maximum height of a projectile is

$H = \frac{v^2\sin^2\theta}{2g}.$

Given that the maximum height is $4R$, so

$\frac{v^2\sin^2\theta}{2g} = 4R.$

3. Substitute $v = \dfrac{2\pi R}{T}$

$\frac{\left(\dfrac{2\pi R}{T}\right)^2 \sin^2\theta}{2g} = 4R.$

Simplify:

$\frac{4\pi^2 R^2}{T^2}\cdot \frac{\sin^2\theta}{2g} = 4R$

$\frac{2\pi^2 R^2}{gT^2}\sin^2\theta = 4R.$

Divide by $R$:

$\frac{2\pi^2 R}{gT^2}\sin^2\theta = 4.$

Hence,

$\sin^2\theta = \frac{4gT^2}{2\pi^2 R} = \frac{2gT^2}{\pi^2 R}.$

So,

$\sin\theta = \left(\frac{2gT^2}{\pi^2 R}\right)^{1/2}.$

Therefore,

$\theta = \sin^{-1}\left(\frac{2gT^2}{\pi^2 R}\right)^{1/2}.$

4. Match with options

This corresponds to Option A:

$\sin ^{-1}\left[\frac{2 gT^2}{\pi^2 R}\right]^{\frac{1}{2}}.$

5. Verification with stored answer

Stored correct answer: A

Derived answer: A

They match.