Question

A particle of mass m is projected with a velocity v at an angle of $60^{\circ }$ with horizontal. When the particle is at its maximum height. The magnitude of its angular momentum about the point of projection is

• A. zero
• B. $\frac{3m{v}^2}{16g}$
• C. $\frac{\sqrt{3}m{v}^3}{16g}$
• D. $\frac{3m{v}^2}{8g}$

Answer 1

Answer: (B)

Maximum height, $H=\frac{v^2{\sin }^{2}60^{\circ }}{2g}$
$=\frac{v^2}{2g}\times \frac{3}{4}=\frac{3v^2}{8g}$
Momentum of particle at highest point $p=mv\cos 60^{\circ }=\frac{mv}{2}$
Angular momentum = pH
$=\frac{mv}{2}\times \frac{3v^2}{8g}$
$=\frac{3m{v}^3}{16g}$