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. along positive z Axis
• B. along negative z Axis
• C. along positive z Axis
• D. along negative z Axis

Answer 1

The correct option is D

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 =

$\therefore$ The required angular momentum

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