Question

A particle of mass '$m$' is projected with velocity '$v$' an angle $\theta$ with the horizontal. Find its angular momentum about the point of projection when it is at the highest point of its trajectory?

• A. $\frac{m{v}^{3}si{n}^2\theta cos\theta }{2g}$ along positive z Axis
• B. $\frac{m{v}^{3}si{n}^2\theta cos\theta }{2g}$along negative z Axis
• C. $\frac{m{v}^{3}si{n}^2\theta cos\theta }{g}$along positive z Axis
• D. $\frac{m{v}^{3}si{n}^2\theta cos\theta }{2g}$along negative z Axis

Answer 1

The correct option is D

$\frac{m{v}^{3}si{n}^2\theta cos\theta }{2g}$along negative z Axis

At the highest point, it has only horizontal velocity

$V_x$ = $vcos\theta$

Now length of the perpendicular to the horizontal velocity from '$O$' is the maximum height reached.

We already know,

$H_{max}=\frac{v^{2}si{n}^2\theta }{2g}$

$\therefore$ The required angular momentum

$L$ = $\frac{m{v}^{3}si{n}^2\theta cos\theta }{2g}$, which is directed along negative $z$ axis.