First slide

Rectilinear Motion

Question

balloon rises from rest with a constant acceleration g/8. A stone is released from it when it has risen to height h. The time taken by the stone to reach the ground is

Moderate

A

$4 \sqrt{\frac{h}{g}}$
B

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

$\sqrt{\frac{2 h}{g}}$
D

$\sqrt{\frac{g}{h}}$

Solution

The velocity of balloon at height h, $v = \sqrt{2 (\frac{g}{8}) h}$
When the stone is released from this balloon, it will go upward with velocity $v = \frac{\sqrt{g h}}{2}$ (same as that of balloon). In this condition, time taken by stone to reach the ground is
$\begin{array}{l} t = \frac{v}{g} [1 + \sqrt{1 + \frac{2 g h}{v^{2}}}] \\ = \frac{\sqrt{g h} / 2}{g} [1 + \sqrt{1 + \frac{2 g h}{g h / 4}}] = \frac{2 \sqrt{g h}}{g} = 2 \sqrt{\frac{h}{g}} \end{array}$

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