Question

A particle of mass $m$ is projected at an angle ${45}^0$ with velocity v . when the particle is at the highest point , the angular momentum with respect with respect to point of projection is

• A. $zero$
• B. $\frac{m{v}^3}{4\sqrt{2}g}$
• C. $\frac{m{v}^2}{\sqrt{2}g}$
• D. $\frac{\sqrt{2}m{v}^3}{4g}$

Answer 1

The correct option is B $\frac{m{v}^3}{4\sqrt{2}g}$

Angular momentum
$J=r\times p$
$=mvcos\theta H$
$=mvcos\theta \frac{v^{2}si{n}^2\theta }{2g}$
at $\theta ={45}^0$
$J=\frac{m{v}^3}{4\sqrt{2}g}$