Question

The speed of a projectile at its maximum height is $\frac{\sqrt{3}}{2}$times its initial speed. If the range of the projectile is P times the maximum height attained by it, then P equals

• A. $\frac{4}{3}$
• B. $\frac{4}{\sqrt{3}}$
• C. $4\sqrt{3}$
• D. $\frac{3}{4}$

Answer 1

Answer: (C)

$4\sqrt{3}$

Answer 2

Let u be initial speed and $\theta$is the angle of the projections.

As per questions Speed at the maximum height

$v_{H} = u\cos\theta = \frac{\sqrt{3}}{2}u$ $\therefore\cos\theta = \frac{\sqrt{3}}{2}$

$\theta = \cos^{- 1}\left( \frac{\sqrt{3}}{2} \right) = 30{^\circ}$

Range, $R = \frac{u^{2}\sin 2\theta}{g}$

Maximum height, $H = \frac{u^{2}\sin^{2}\theta}{2g}$

As R = PH (given)

$\therefore\frac{u^{2}\sin 2\theta}{g} = p\frac{u^{2}\sin^{2}\theta}{2g}$

$2\sin\theta\cos\theta = \frac{P}{2}\sin^{2}\theta$

$\tan\theta = \frac{4}{P}$

Or $P = \frac{4}{\tan 30{^\circ}} = 4\sqrt{3}$