Q. A particle is projected at $60{^\circ}$ to the horizontal with a kinetic energy$\ K$. The kinetic energy at the highest point is
A) $K$
B) $Zero\ $
C) $\frac{K}{4}$
D) $\frac{K}{2}$
Explanation
(c) Kinetic energy at highest point
$(KE)_{H} = \frac{1}{2}mv^{2}\cos 2\theta$
$= K\cos^{2}\theta$
$= K({\cos{60{^\circ})}}^{2}$
$= \frac{K}{4}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.