Question

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

Answer 1

As shown in figure , Just strikes the ground $u_x$ = $ucos\left(\theta \right)$ , $u_y=usin\left(\theta \right)$,

Time of flight = $\frac{2usin\left(\theta \right)}{g}$,

$\therefore$ Range $R=ucos\left(\theta \right)\times$ $\frac{2usin\left(\theta \right)}{g}$,

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

Angular Momentum $\overline{L}=r\times mu$,

Here $r=R=\frac{u^{2}sin\left(2\theta \right)}{g}\hat{i}$, $mu=m\left(ucos\right(\theta )\hat{i}-usin(\theta \left)\hat{j}\right)$

$\therefore \overline{L}=\frac{m{u}^{3}sin\left(2\theta \right)}{g}\times \left(cos\right(\theta )\hat{i}\times \hat{i}-sin(\theta )\hat{i}\times \hat{j})$

$\Rightarrow \overline{L}=\frac{m{u}^{3}sin\left(\theta \right)sin\left(2\theta \right)}{g}(-\hat{k})$