Question

A particle of mass $m$ is projected with a velocity $\upsilon$ 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. zero
• B. $\frac{m\upsilon^{3}}{\sqrt{2}g}$
• C. $\frac{\sqrt{3}}{16}\frac{ m\upsilon^{3}}{g}$
• D. $\frac{\sqrt{3}}{2}\frac{ m\upsilon^{2}}{g}$

Answer 1

Answer: (C)

$\frac{\sqrt{3}}{16}\frac{ m\upsilon^{3}}{g}$

Answer 2

$L_{0} = Pr_{\bot}$ $L_{0} = mv\, cos\theta\, H$ $= mg \frac{\sqrt{3}}{2}. \frac{v^{2}\,sin^{2}\,30^{\circ}}{2g}\quad= \quad\frac{ \sqrt{3}m\upsilon^{3}}{16g}$