QJEEM24Ph9423: JEE Main 2024 Physics - Class 11 JEEM24Ph9423 Solution: A particle moving in a circle of radius with uniform speed…

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.

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:

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

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 $v_y=v\sin\theta$ .

The maximum height $H$ reached by the projectile can be found from the kinematic equation:

$H = \frac{v_y^2}{2g} = \frac{(v\sin\theta)^2}{2g}$

We are given that the maximum height attained $H$ is equal to $4R$ , so:

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

Substitute $v$ from the first equation into the second one:

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

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

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

$2gT^2 = \pi^2 R \sin^2\theta$

Now solve for $\sin\theta$ :

$\sin\theta =\left[\frac{2gT^2}{\pi^2 R}\right]^{1/2}$

Finally, to solve for $\theta$ , take the inverse sine of both sides:

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

Therefore, the correct answer is:

Option A

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