Q. A projectile is projected with kinetic energy $K$. If it has the maximum possible horizontal range, then its kinetic energy at the highest point will be
A) $0.25\ K$
B) $0.5\ K$
C) $0.75\ K$
D) $1.0\ K$
Explanation
(b) Since range is max, therefore $\theta = 45{^\circ}$
Hence, $V_{x} = V\cos\theta = V\cos{45{^\circ}} = \frac{V}{\sqrt{2}}$
At the highest point, the net velocity of the projectile is
$V_{x} = V\cos{45{^\circ}}$
$\therefore K.E. = \frac{1}{2}mV_{x}^{2} = \frac{1}{2}m\frac{V^{2}}{2} = 0.5\ K$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.