Question

A ball of mass ‘m’ is dropped from a height ‘h’ above the ground. Calculate speed of the ball just before hitting the ground.

• A. 2gh
• B. $\sqrt{2gh}$
• C. $\sqrt{gh}$
• D. gh

Answer 1

The correct option is B $\sqrt{2gh}$

Mass of the ball = m
height = h
Potential energy at height h, (P.E.) = mgh
Kinetic enercy just before hitting the ground (K.E.) $=\frac{1}{2}m{v}^2$
P.E. become zero just before hitting the ground.
Hence This P.E. converted into K.E. ,
So both must be equal P.E. = K.E.
$mgh=\frac{1}{2}m{v}^2$
$v^2=2gh$
$v=\sqrt{2gh}$