Q. A stone of mass 2 kg is projected upward with KE of 98 J. The height at which the KE of the body becomes half its original value, is given by (Take $g = 10ms^{- 2}$)
A) 5 m
B) 2.5 m
C) 1.5 m
D) 0.5 m
Explanation
(b) $K = \frac{1}{2}mv^{2}$
$v^{2} = \frac{98 \times 2}{2} = 98$
$h = \frac{v^{2}}{2g} = \frac{98}{2 \times 9.8} = 5$
$K_{1} = \frac{1}{2}mv^{2} = \frac{1}{2}m \times 2gh$
$\therefore\ \ \ \ \ \frac{K_{2}}{K_{1}} = \frac{h_{2}}{h_{1}}$
Given $K_{2} = \frac{K_{1}}{2}$
$\therefore\ \ \ = \frac{K_{1}}{2K_{1}} = \frac{h_{2}}{5}$
$\therefore\ h_{2} = 2.5\ m$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.