Question

A ball of mass $m$ is projected with velocity $u$ making an angle $\theta$ with the horizontal. What is its angular momentum at the highest point?

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

Answer 1

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

$L=m\times v\times r$
$r=$perpendicular distance
Height of the projectile $=H_{\text{max}}=\frac{u^2{\sin }^2\theta }{2g}$
Velocity at heighest point $V_{HZ}=u\cos \theta$
$L=m\times \left(u\cos \theta \right)\times \left(\frac{u^2{\sin }^2\theta }{2g}\right)$