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

The Correct Option is A

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:

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