Question

A particle is projected with a velocity $v$ so that its horizontal range is twice the maximum height attained. The horizontal range is

• A. $\frac{v^2}{g}$
• B. $\frac{2v^2}{3g}$
• C. $\frac{4v^2}{5g}$
• D. none of these

Answer 1

The correct option is C $\frac{4v^2}{5g}$

Given,

Range, $R=2H$

But, $R=4Hcot\theta$

$⟹cot\theta =\frac{1}{2}$

From triangle,

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

So, the range of the projectile,

$R=\frac{2v^{2}sin\theta cos\theta }{g}=\frac{2v^2}{g}\times \frac{2}{\sqrt{5}}\times \frac{1}{\sqrt{5}}$

$=\frac{4v^2}{5g}$