Q. A ball dropped from a height of $2m$ rebounds to a height of $1.5\ m$ after hitting the ground. Then the percentage of energy lost is
A) 25
B) 30
C) 50
D) 100
Explanation
(a) $U_{1} = mgh_{1}$ and $U_{2} = mgh_{2}$
% energy lost $= \frac{U_{1} - U_{2}}{U_{1}} \times 100$
$= \frac{mgh_{1} - mgh_{2}}{mgh_{1}}100 = \left( \frac{h_{1} - h_{2}}{h_{1}} \right) \times 100$
$= \frac{2 - 1.5}{2} \times 100 = 25\%$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.