First slide

Projectile motion

Question

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

Moderate

A

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

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

$\frac{4 v^{3}}{5 g^{2}}$
D

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

Solution

$Since R = 2 H$
or $\frac{v^{2} \sin 2 θ}{g} = 2 \times \frac{v^{2} \sin^{2} θ}{2 g}$
$\begin{array}{l} or 2 \sin θ \cos θ = \sin^{2} θ or \tan θ = 2 \\ R = \frac{v^{2} \sin 2 θ}{g} \\ = \frac{v^{2} 2 \sin θ \cos θ}{g} = \frac{2 v^{2}}{g} \times \frac{2}{\sqrt{5}} \times \frac{1}{\sqrt{5}} = \frac{4 v^{2}}{5 g} \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App