Q. A man throws a piece of stone to a height of 12 m where it reaches with a speed of $12\ ms^{- 1}$. If he throws the same stone such that it just reaches this height, the percentage of energy saved is nearly
A) 19%
B) 38%
C) 57%
D) 76%
Explanation
(b) Percentage of energy saved is
$\frac{\frac{1}{2}mv^{2} \times 100}{\frac{1}{2}mv^{2} + mgh} = \frac{v^{2} \times 100}{v^{2} + 2gh}$
$ie,\ = \frac{12 \times 12 \times 100}{12 \times 12 + 2 \times 10 \times 12}$
$= \frac{14400}{144 + 240} = \frac{14400}{384} = 37.5$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.