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A particle of mass ' $m$ ' is projected with a velocity $v$ making an angle of $30^{\circ}$ with the horizontal. The magnitude of angular momentum of the projectile about the point of projection when the particle is at its maximum height ' $h$ ' is
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$\frac{\sqrt{3}}{16} \frac{m v^{3}}{g}$
Angular momentum $L_{0}=p r_{\perp}$ $\left(\because\right.$ linear momentum $p=m v \cos \theta$ and $\left.\mathrm{r}_{1}=H\right)$
$\Rightarrow L_{0}=m v \cos \theta H$
$=m v \frac{\sqrt{3}}{2} \cdot \frac{v^{2} \sin ^{2} 30^{\circ}}{2 g}$
$=\frac{\sqrt{3} m v^{3}}{16 g}$
$\Rightarrow L_{0}=m v \cos \theta H$
$=m v \frac{\sqrt{3}}{2} \cdot \frac{v^{2} \sin ^{2} 30^{\circ}}{2 g}$
$=\frac{\sqrt{3} m v^{3}}{16 g}$
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