Question

A body is projected with a velocity $u$ so that its horizontal range is twice the greatest height attained. The value of its range is
(Take $g=10{\text{m/s}}^2$)

• A. $\frac{u^2}{10}$
• B. $\frac{u^2}{25}$
• C. $\frac{2u^2}{25}$
• D. $\frac{u^2}{50}$

Answer 1

The correct option is C $\frac{2u^2}{25}$

By using the relation
$4H=R\tan \theta$
According to question, $R=2H$
$\Rightarrow 4H=\left(2H\right)\tan \theta$
$\Rightarrow \tan \theta =2$
$\Rightarrow \sin \theta =\frac{2}{\sqrt{5}},\cos \theta =\frac{1}{\sqrt{5}}$
Range of the projectile is
$R=\frac{u^2\sin 2\theta }{g}$
$R=\frac{u^2(2\times \sin \theta \cos \theta )}{g}$
$R=\frac{u^2(2\times \frac{2}{\sqrt{5}}\times \frac{1}{\sqrt{5}})}{10}=\frac{2u^2}{25}$