Question

If R and H represent horizontal range and maximum height of the projectile, then the angle of projection with the horizontal is

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

Answer 1

Answer: (C)

$\tan^{- 1}\left( \frac{4H}{R} \right)$

Answer 2

Maximum height, $H = \frac{u^{2}\sin^{2}\theta}{2g}$ ….. (i)

Horizontal range, $R = \frac{u^{2}\sin 2\theta}{g} = \frac{2u^{2}\sin\theta\cos\theta}{g}$..(ii)

Divide (i) by (ii), we get

$\frac{H}{R} = \frac{\tan\theta}{4}$or$\tan\theta = \frac{4H}{R}or\theta = \tan^{- 1}\left( \frac{4H}{R} \right)$