First slide

Projection Under uniform Acceleration

Question

If R and H represent horizontal range and maximum height of the projectile, then the angle of projection with the horizontal is

Moderate

A

$\tan^{- 1} (\frac{H}{R})$
B

$\tan^{- 1} (\frac{2 H}{R})$
C

$\tan^{- 1} (\frac{4 H}{R})$
D

$\tan^{- 1} (\frac{4 R}{H})$

Solution

Maximum height, H = $\frac{u^{2} \sin^{2} θ}{2 g} - - - - - - (i)$
Horizontal range, R = $\frac{u^{2} \sin 2 θ}{g} = \frac{2 u^{2} \sin θ \cos θ}{g}$ -------------(ii)
Dividing (i) by (ii), we get
$\frac{H}{R} = \frac{\tan θ}{4} or \tan θ = \frac{4 H}{R} or θ = \tan^{- 1} (\frac{4 H}{R})$

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