Question

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

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

Answer 1

The correct option is A

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

$R=\frac{V^{2}sin2\theta }{g}=2\times \frac{V^{2}sin\theta }{g}$

$H=\left(V^{2}si{n}^2\frac{\theta }{2g}\right)$

Given, $R=2H$

$2sin\theta cos\theta =si{n}^2\theta$
$tan\theta =2$
$sin\theta =\frac{2}{\sqrt{5}}cos\theta =\frac{1}{\sqrt{5}}$
$R=\frac{V^{2}sin2\theta }{g}=V^2\times \frac{2sin\theta cos\theta }{g}=V^2\times \frac{2}{g}\times \frac{2}{\sqrt{5}}\times \frac{1}{\sqrt{5}}$
$R=\frac{4V^2}{5g}$