Q. A particle of mass $m\ $is projected with a velocity $v$ making an angle of $30{^\circ}$ 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) $\frac{\sqrt{3}}{2}\frac{{mv\ }^{2}}{g}$
B) $Zero$
C) $\frac{{mv}^{3}}{\sqrt{2g}}$
D) $\frac{\sqrt{3}}{16}\frac{{mv}^{3}}{g}$
Explanation
(d) Angular momentum of the projectile
$L = mv_{h}r_{\bot}$
$= m(vcos\theta)h$
(where $h$ is the maximum height)
$\Longrightarrow \ = m(vcos\theta)\left( \frac{v^{2}\sin^{2}\theta}{2g} \right)$
$L = \frac{{mv}^{3}\sin^{2}\theta cos\theta}{2g} = \frac{\sqrt{3}{mv}^{3}}{16g}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.