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. $2\sqrt{3}$
• C. $4\sqrt{3}$
• D. $\frac{3}{4}$

Answer 1

Answer: (C)

Let $u$ be the initial speed and $\theta$ is the angle of the projection.
Speed at the maximum height,
$v_H=ucos\theta =\frac{\sqrt{3}}{2}u$
$\therefore cos\theta =\frac{\sqrt{3}}{2}$ or
$\theta =co{s}^{-1}\left(\frac{\sqrt{3}}{2}\right)=30^{\circ }$
Range,
$R=\frac{u^{2}sin2\theta }{g}$ and maximum height,
$H=\frac{u^{2}si{n}^2\theta }{2g}$
As $R=PH$ (Given)
$\therefore \frac{u^{2}sin2\theta }{g}=P\frac{u^{2}si{n}^2\theta }{2g}$
$2sin\theta cos\theta =\frac{P}{2}si{n}^2\theta$ or
$tan\theta =\frac{4}{P}$
$\therefore P=\frac{4}{tan30^{\circ }}=4\sqrt{3}$