Question

A shell of mass $m$ is projected with a velocity $v$ at an angle ${60}^o$ to horizontal. When it reaches the maximum height, its angular momentum with respect to point of projection is

• A. $\frac{\sqrt{3}{mv}^3}{8g}$
• B. $\frac{3{mv}^3}{8g}$
• C. $\frac{3{mv}^3}{16g}$
• D. $\frac{\sqrt{3}{mv}^3}{16g}$

Answer 1

The correct option is D $\frac{\sqrt{3}{mv}^3}{16g}$

Vertical Velocity, $V_V=vsin\left({60}^o\right)=\frac{v}{2}$

Horizontal velocity, $V_H=vcos\left({60}^o\right)=\frac{\sqrt{3}v}{2}$

Let maximum height is $h$,

Using $v^2=u^2+2as$ $\Rightarrow 0=(\frac{v}{2}{)}^2-2gh$

$\Rightarrow h=\frac{v^2}{8g}$

At highest point , Velocity$=V_H$,

Angular Momentum$=r\times mv=\frac{v^2}{8g}\times \frac{\sqrt{3}mv}{2}=\frac{\sqrt{3}m{v}^3}{16g}$