First slide

Projection Under uniform Acceleration

Question

A stone is thrown at an angle e to the horizontal reaches a maximum height h. Then the time of flight of stone will be

Easy

A

$\sqrt{(\frac{2 h}{g})}$
B

$2 \sqrt{(\frac{2 h}{g})}$
C

$\frac{2 \sqrt{2 \sin θ}}{g}$
D

$\frac{2 \sqrt{2 hsin θ}}{g}$

Solution

We know that, $h = u^{2} \sin^{2} θ / 2 g$
$∴ usin θ = \sqrt{(2 gh)}$
Now, $T = \frac{2 usin θ}{g} = \frac{2 \times \sqrt{(2 gh)}}{g}$
$= 2 \sqrt{(\frac{2 h}{g})}$

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