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

Answer 1

Answer: (D)

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