Question

An aeroplane is moving horizontally with a velocity $u$. It drops a packet when it is at height $h$. The time taken by the packet in reaching the ground will be

• A. $\sqrt{\frac{2h}{g}}$
• B. $\sqrt{\frac{2u}{g}}$
• C. $\sqrt{\frac{h}{2g}}$
• D. $\sqrt{\frac{2g}{h}}$

Answer 1

Answer: (A)

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

The horizontal velocity of the aeroplane or the packet does not play a role in the time taken for the packet to reach the ground.

Using the kinematic equation,

$h=ut+\frac{1}{2}g{t}^2$.

Initial vertical velocity of the packet in the downward direction $u=0$.

$\therefore h=0+\frac{1}{2}g{t}^2$

$\therefore t=\sqrt{\frac{2h}{g}}$