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A particle of mass m is projected with a velocity v making an angle of 45^o with the horizontal. The magnitude of angular momentum of the projectile about an axis of projection when the particle is at maximum height h is

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a

0

b

${mv}^{3} / 4 \sqrt{2} g$

c

${mv}^{2} / \sqrt{2} g$

d

$m \sqrt{(2 {gh}^{3})}$

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Detailed Solution

$0 - v^{2} \sin^{2} 45^{\circ} = 2 gh or h = v^{2} / 4 g$
At highest point, momentum
$= mvcos 45^{\circ} = mv / \sqrt{2}$
Angular momentum
$= \frac{mv}{\sqrt{2}} \times h = \frac{mv}{\sqrt{2}} \times \frac{v^{2}}{4 g} = \frac{{mv}^{3}}{4 \sqrt{2} g}$
In terms of h, angular momentum
$= \frac{mv}{\sqrt{2}} \times h = \frac{m}{\sqrt{2}} [\sqrt{(4 gh)}] \times h = m \sqrt{(2 {gh}^{3})}$

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