Question

A particle of mass $m$ is projected with velocity $v$ at 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. $\frac{m{v}^2{\sin }^2\theta \cos \theta }{2g}$
• B. $\frac{m{v}^3\sin \theta \cos \theta }{g}$
• C. $\frac{m{v}^3{\sin }^2\theta \cos \theta }{2g}$
• D. $\frac{m{v}^3{\sin }^2\theta \cos \theta }{g}$

Answer 1

The correct option is C $\frac{m{v}^3{\sin }^2\theta \cos \theta }{2g}$

At the highest point, it has horizontal velocity only i.e. $v_x=v\cos \theta$.
Length of the perpendicular to the horizontal velocity from ${}^{\prime }{O}^{\prime }$ is the maximum height, where
$H_{max}=\frac{v^2{\sin }^2\theta }{2g}$

We know,
$L=m{r}_{\perp }v$
$=m\times H_{max}\times v\cos \theta$
$=m\times \left(\frac{v^2{\sin }^2\theta }{2g}\right)\times v\cos \theta$
$=\frac{m{v}^3{\sin }^2\theta \cos \theta }{2g}$
$\therefore$ Option (c) is correct.