A particle moving in a circle of radius R with a uniform speed takes a time T to complete one revolution. If this particle were projected with the same speed at an angle ‘θ’ to the horizontal, the maximum height attained by it equals 4R. The angle of projection, θ, is then given by :

The correct option/answer is given by Option 3, which corresponds to:

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

Step-by-step Derivation:

1. Uniform Circular Motion of the Particle:
The particle moves in a circle of radius R with a uniform speed v. It takes time T to complete one full revolution (distance equal to the circumference $2\pi R$).
Therefore, the speed of the particle is given by:

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

2. Projectile Motion of the Particle:
The particle is projected with the same speed v at an angle $\theta$ to the horizontal. The formula for the maximum height H attained by a projectile is:

$H=\frac{v^2{\sin }^2\theta }{2g}$
Here, g is the acceleration due to gravity.

3. Equating the Maximum Height to 4R:
We are given that the maximum height attained by the particle equals 4R:

$H=4R$
Substitute the formula for H:

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

4. Substituting the Value of Speed (v):
Substitute $v=\frac{2\pi R}{T}$ into the height equation:

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

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

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

5. Solving for $\theta$:
Divide both sides by 2R:

$\frac{{\pi }^{2}R{\sin }^2\theta }{g{T}^2}=2$
Rearranging terms to solve for ${\sin }^2\theta$:

${\sin }^2\theta =\frac{2g{T}^2}{{\pi }^{2}R}$
Taking the square root on both sides:

$\sin \theta ={\left[\frac{2g{T}^2}{{\pi }^{2}R}\right]}^{1/2}$
Thus, the angle of projection $\theta$ is:

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