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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
zero
$\frac{\sqrt{3}}{16} \cdot \frac{ mv ^{3}}{ g }$
$\frac{{m{v^3}}}{{\sqrt 2 g}}$
$\frac{\sqrt{3}}{2} \cdot \frac{ mv ^{2}}{ g }$
Solution
Angular momentum, $\overrightarrow{ L }=\overrightarrow{ r } \times m \overrightarrow{ v }$
$\Rightarrow|\overrightarrow{ L }|= rmv \sin \theta$
At maximum point, velocity is $v = v \cos \theta= v \cos (30)=\frac{\sqrt{3} v }{2}$ only (direction: towards horizontal) and no vertical velocity is present.
The maximum height reached will be
$h=\frac{v^{2} \sin ^{2} \theta}{2 g}$
$\Rightarrow h =\frac{ v ^{2} \sin ^{2}\left(30^{\circ}\right)}{2 g }$
$\Rightarrow h =\frac{ v ^{2}}{8 g }$
$L = rmv \sin \theta$
$\Rightarrow L = mvh$
$\Rightarrow L = m \times \frac{\sqrt{3} v }{2} \times \frac{ v ^{2}}{8 g }$
$\Rightarrow L =\frac{\sqrt{3} mv ^{3}}{16 g }$
