First slide

Projection Under uniform Acceleration

Question

A particle is projected with a velocity $ν$ , so that its range on a horizontal plane is twice the greatest height attained. If 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

$H = \frac{v^{2} \sin^{2} θ}{2 g} and R = \frac{v^{2} \sin 2 θ}{g}$
Since, $R = 2 H, so \frac{v^{2} \sin 2 θ}{g} = 2 \frac{v^{2} \sin^{2} θ}{2 g}$
or $2 \sin θcos θ = \sin^{2} θ or \tan θ = 2$
$\begin{array}{l} ∴ R = v^{2} \frac{2}{g} \sin θcos θ \\ = \frac{2 v^{2}}{g} \frac{2}{\sqrt{5}} \frac{1}{\sqrt{5}} = \frac{4 v^{2}}{5 g} \end{array}$

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