Question

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. $=\frac{\sqrt{3}}{2}\frac{m{v}^2}{g}$
• B. $Zero$
• C. $\frac{m{v}^3}{\sqrt{2}g}$
• D. $=\frac{\sqrt{3}}{16}\frac{m{v}^3}{g}$

Answer 1

The correct option is D $=\frac{\sqrt{3}}{16}\frac{m{v}^3}{g}$

Angular momentum of the projectile = m v r
$=m.\left(vcos\theta \right).h=m\left(vcos\theta \right)\frac{v^{2}si{n}^2\theta }{2g}=\frac{m{v}^{3}si{n}^2\theta cos\theta }{2g}=\frac{\sqrt{3}}{16}\frac{m{v}^3}{g}$