A particle of mass m is projected with velocity V making an angle of… | MHT CET 2026 (14 May, Shift 2)

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Angular Momentum & its Conservation

A particle of mass m is projected with velocity V making an angle of $45^\circ$ with the horizontal. The magnitude of the angular momentum of the projectile about an axis of projection when the particle is at maximum height h is ( $g$ = acceleration due to gravity) $\left(\sin 45^\circ = \frac{1}{\sqrt{2}}\right)\ \left(\cos 45^\circ = \frac{1}{\sqrt{2}}\right)$

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Answer: (b)

At maximum height the velocity is horizontal: $V\cos 45^\circ = \dfrac{V}{\sqrt{2}}$ ; perpendicular distance from the point of projection is $h$ .
$h = \dfrac{V^2\sin^2 45^\circ}{2g} = \dfrac{V^2}{4g}$
$L = m\dfrac{V}{\sqrt{2}}\cdot\dfrac{V^2}{4g} = \dfrac{mV^3}{4\sqrt{2}g}$
So the answer is (b).

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