Question

A particle moving in a circle of radius R with uniform speed takes time T to complete one revolution. If this particle is projected with the same speed at an angle θ to the horizontal, the maximum height attained by it is equal to 4R. The angle of projection θ is then given by

• A. $\sin^{-1} \left[ \sqrt{\frac{2gT^2}{\pi^2R}} \right]$
• B. $\sin^{-1} \left[ \sqrt{\frac{\pi^2R}{2gT^2}} \right]$
• C. $\cos^{-1} \left[ \sqrt{\frac{2gT^2}{\pi^2R}} \right]$
• D. $\cos^{-1} \left[ \sqrt{\frac{\pi R}{2gT^2}} \right]$

Answer 1

Answer: (A)

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

Answer 2

Given:

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

The maximum height attained by the particle is given by:

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

We are given that:

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

Simplifying:

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

Taking the square root:

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

Thus:

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