Q. A particle is projected with a speed $v$ at $45{^\circ}$ with the horizontal. The magnitude of angular momentum of the projectile about the point of projection when the particle is at its maximum height $h$ is
A) Zero
B) $\frac{mvh^{2}}{\sqrt{2}}$
C) $\frac{mvh}{\sqrt{2}}$
D) $\frac{mvh^{3}}{\sqrt{2}}$
Explanation
(c) Velocity of particle at maximum height $h$ is $v^{'} = v\cos\theta$ where $v =$ initial velocity of particle at which it is projected, $\theta =$ angle of projection
Angular momentum, $L = mv^{'}h = mv\cos\theta h$
$= mvh\cos{45{^\circ}} = \frac{mvh}{\sqrt{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.