Question

A particle of mass $m$ is projected from horizontal ground with an initial velocity of $u$ making an angle $\theta$ with horizontal. The angular momentum of the particle about the point of projection, when it just strikes the ground, is -

• A. $0$
• B. $\frac{m{u}^3{\sin }^2\theta \cos \theta }{2g}$
• C. $\frac{m{u}^3{\sin }^2\theta {\cos }^2\theta }{g}$
• D. $\frac{2m{u}^3{\sin }^2\theta \cos \theta }{g}$

Answer 1

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


Angular momentum about point $A$ when the particle hits ground at $B$ is,

$L=muR\sin \theta$

$=mu\left(\frac{u^2\sin 2\theta }{g}\right)\sin \theta$


$=\frac{2m{u}^3{\sin }^2\theta \cos \theta }{g}$

Hence, option $\left(D\right)$ is the correct answer.