Question

A particle of mass $m$ is projected with velocity $u$ at an angle of $\theta$ with the horizontal. The initial angular momentum of the particle about the highest point of its trajectory is equal to

• A. $\frac{m{u}^{3}si{n}^2\theta cos\theta }{3g}$
• B. $\frac{3m{u}^{3}si{n}^2\theta cos\theta }{2g}$
• C. $\frac{m{u}^{3}si{n}^2\theta cos\theta }{2g}$
• D. None of these

Answer 1

The correct option is C $\frac{m{u}^{3}si{n}^2\theta cos\theta }{2g}$

Taking the direction into the plane of the paper as positive,
$L_P=\left(musin\theta \right)\left(\frac{R}{2}\right)-\left(mucos\theta \right)\left(H\right)$

$=\left(musin\theta \right)\left(\frac{usin\theta }{g}\right)\left(ucos\theta \right)-\left(mucos\theta \right)\left(\frac{u^{2}si{n}^2\theta }{2g}\right)$
$=\frac{m{u}^{3}si{n}^2\theta cos\theta }{2g}$.