Q. A body is projected at an angle $\theta$ to the horizontal with kinetic energy $E_{k}$. The potential energy at the highest point of the trajectory is
A) $E_{k}$
B) $E_{k}\cos^{2}\theta$
C) $E_{k}\sin^{2}\theta$
D) $E_{k}\tan^{2}\theta$
Explanation
(c) Let $v$ be the velocity of projection and $\theta$ the angle of projection
Kinetic energy at highest point
$= \frac{1}{2}mv^{2}\cos^{2}\theta\ or\ \ E_{k}\cos^{2}\theta$
Potential energy at highest point
$= E_{k} - E_{k}\cos^{2}\theta = E_{k}\left( 1 - \cos^{2}\theta \right) = E_{k}\sin^{2}\theta$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.