Q. A particle is projected with a speed $v$ at $45^{0}$ with the horizontal. The magnitude of angular momentum of the projectille about the point of projection when the particle is at its maximum height $h$ is
A) Zero
B) $\frac{{mvh}^{2}}{\sqrt{2}}$
C) $\frac{mv^{2}h}{\sqrt{2}}$
D) $\frac{{mv}^{2}h}{\sqrt{2}}$
Explanation
(d) When a particle is projected with a speed $v$ at $45{^\circ}$ with the horizontal then velocity of the projectile at maximum height.
$v^{'} = v\cos{45{^\circ} = \frac{v}{\sqrt{2}}}$
Angular momentum of the projectile about the point of projection
$= mv'h$
$= m\frac{v}{\sqrt{2}}h = \frac{mvh}{\sqrt{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.