Question

The maximum height of a projectile is half of its range on the horizontal. If the velocity of projection is $u,$ its range on the horizontal is

• A. $\frac{2 u^{2}}{5 g}$
• B. $\frac{3 u^{2}}{5 g}$
• C. $\frac{u^{2}}{g}$
• D. $\frac{4 u^{2}}{5 g}$

Answer 1

Answer: (D)

Maximum height, $H_{\max }=\frac{u^{2} \sin ^{2} \theta}{2 g}$
Horizontal range, $R=\frac{u^{2} \sin 2 \theta}{g}$
According to the given problem,
$H_{\max }=\frac{R}{2}$
$\therefore \frac{u^{2} \sin ^{2} \theta}{2 g}=\frac{u^{2} \sin 2 \theta}{2 g}$
or $\frac{u^{2} \sin ^{2} \theta}{2 g}=\frac{2 u^{2} \sin \theta \cos \theta}{2 g}$
or $\tan \theta=2$
$\therefore \sin \theta=\frac{2}{\sqrt{5}} ; \cos \theta=\frac{1}{\sqrt{5}}$
$\therefore R=\frac{2 u^{2} \sin \theta \cos \theta}{g}$
$=\frac{2 u^{2} \times \frac{2}{\sqrt{5}} \times \frac{1}{\sqrt{5}}}{g}$
$=\frac{4 u^{2}}{5 g}$