A particle is projected with a speed v at 45 ° with the horizontal. The magnitude of angular momentum of the projectille about the point of projection when the particle is at its maximum height h is

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When a particle is projected with a speed $v$ at $45 °$ with the horizontal then velocity of the projectile at maximum height.

$v^{'} = v \cos 45 ° = \frac{v}{\sqrt{2}}$

Angular momentum of the projectile about the point of projection

$= m v' h$

$= m \frac{v}{\sqrt{2}} h = \frac{m v h}{\sqrt{2}}$

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