JEE Main 2024 — Circular Motion Question with Solution

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:

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

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

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

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

Now solve for $\sin \theta$:

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

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

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

Therefore, the correct answer is:

Option A

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