Q. A body of mass 5 kg is thrown vertically up with a kinetic energy of 490 J. The height at which the kinetic energy of the body becomes half of the original value is
A) 12.5m
B) 10m
C) 2.5m
D) 5m
Explanation
(d) According to law of conservation of energy
$\frac{1}{2}mu^{2} = \frac{1}{2}mv^{2} + mgh$
$490 = 245 + 5 \times 9.8 \times h$
$h = \frac{245}{49} = 5m$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.