A particle of mass $\mathrm{m}$ is projected with a velocity '$\mathrm{u}$' 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 $\mathrm{h}$ is :

Question

$\frac{\mathrm{mu}^3}{\sqrt{2} \mathrm{~g}}$

zero

$\frac{\sqrt{3}}{2} \frac{\mathrm{mu}^2}{\mathrm{~g}}$

$\frac{\sqrt{3}}{16} \frac{\mathrm{mu}^3}{\mathrm{~g}}$

Solution

Please login to view the detailed solution steps...

Go to DASH