Q. An aeroplane is moving with horizontal velocity $u$ at height $h.$The velocity of a packet dropped from it on the earth’s surface will be ($g$ is acceleration due to gravity)
A) $\sqrt{u^{2} + 2gh}$
B) $\sqrt{2gh}$
C) $2gh$
D) $\sqrt{u^{2} - 2gh}$
Explanation
(a) Horizontal velocity of dropped packet $= u$
Vertical velocity $= \sqrt{2gh}$
$\therefore Resultant\ velocity\ at\ earth = \sqrt{u^{2} + 2gh}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.