Question

A projectile of mass $m$ is fired with velocity $u$ making angle $\theta$ with the horizontal. Its angular momentum about the point of projection when it hits the ground is given by

• A. $\frac{2mu{\sin }^2\theta \cos \theta }{g}$
• B. $\frac{2m{u}^3{\sin }^2\theta \cos \theta }{g}$
• C. $\frac{mu{\sin }^2\theta \cos \theta }{2g}$
• D. $\frac{m{u}^3{\sin }^2\theta \cos \theta }{2g}$

Answer 1

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

$L=\overrightarrow{R}\times \overrightarrow{p}$
Where $Range,R=\frac{u^2\sin 2\theta }{g}$
The angle between $\overrightarrow{R}$ and $\overrightarrow{p}=\theta$
Also, $p=mu$.
Hence , $L=\frac{u^2\sin 2\theta }{g}\times mu\sin \theta$
$L=\frac{2m{u}^3{\sin }^2\theta \cos \theta }{g}$