Question

A particle is projected with a velocity $v$ , so that its range on a horizontal plane is twice the greatest height attained. If $g$ is the acceleration due to gravity, then its range is -

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

Answer 1

Answer: (A)

$\because R=2H$
$\therefore \frac{2 v^{2} s i n \theta c o s \theta }{g}=2\frac{v^{2} s i n^{2} \, \theta }{2 g}\Rightarrow tan\theta =2$
$\Rightarrow R=\frac{2 v^{2} \left(s i n \theta \right) \left(\right. c o s \theta \left.\right)}{g}=\frac{2 v^{2} \left(\frac{2}{\sqrt{5}} \, \right) \left(\frac{1}{\sqrt{5}}\right)}{g}=\frac{4 v^{2}}{5 g}$