Question

A body is projected with velocity $u$ so that its horizontal range is twice the greatest height attained. The value of range is

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

Answer 1

The correct option is B $\frac{4u^2}{5g}$

Let's consider, $u=$ initial velocity of the projectile

$\theta =$ angle of projection.

According to question,

$R=2H$

$\frac{u^{2}sin2\theta }{g}=2\frac{u^{2}si{n}^2\theta }{2g}$

$2sin\theta .cos\theta =sin\theta .sin\theta$

$tan\theta =2$

$sin\theta =\frac{2}{\sqrt{5}}$, $cos\theta =\frac{1}{\sqrt{5}}$

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

$R=\frac{2u^2}{g}\frac{2}{\sqrt{5}}.\frac{1}{\sqrt{5}}$

$R=\frac{4u^2}{5g}$

The correct option is B.