Q. A ball is projected with kinetic energy $E$ at an angle of $45{^\circ}$ to the horizontal. At the highest point during its flight, its kinetic energy will be
A) Zero
B) $\frac{E}{2}$
C) $\frac{E}{\sqrt{2}}$
D) $E$
Explanation
(b) New kinetic energy $E^{'} = E = \cos^{2}\theta$
$= E\cos^{2}(45{^\circ})$
$\frac{E}{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.