Question

The angle of projection with the horizontal in terms of maximum height attained and horizontal range is given by

• A. \( \tan^{-1} \left( \frac{2H}{3R} \right) \)
• B. \( \tan^{-1} \left( \frac{4R}{3H} \right) \)
• C. \( \tan^{-1} \left( \frac{4H}{R} \right) \)
• D. \( \tan^{-1} \left( \frac{R}{H} \right) \)

Hint:

Use identities like \( \sin 2\theta = 2 \sin \theta \cos \theta \) to simplify ratios in projectile motion equations.

Answer 1

The Correct Option is C

From projectile motion: \[ H = \frac{u^2 \sin^2 \theta}{2g}, \quad R = \frac{u^2 \sin 2\theta}{g} \] We can write: \[ \frac{H}{R} = \frac{u^2 \sin^2 \theta / 2g}{u^2 \sin 2\theta / g} = \frac{\sin^2 \theta}{2 \sin 2\theta} \] But \( \sin 2\theta = 2 \sin \theta \cos \theta \), so: \[ \frac{H}{R} = \frac{\sin^2 \theta}{4 \sin \theta \cos \theta} = \frac{\sin \theta}{4 \cos \theta} = \frac{1}{4} \tan \theta \Rightarrow \tan \theta = \frac{4H}{R} \Rightarrow \theta = \tan^{-1} \left( \frac{4H}{R} \right) \]