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) $E/2$
C) $E/\sqrt{2}$
D) $E$
Explanation
(b) $E^{'} = E\cos^{2}\theta = E\cos^{2}{(45{^\circ})} = \frac{E}{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.