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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Aicharya · Accepted Answer

When a particle is projected with a speed $v$ at $45^{\circ}$ with the horizontal then velocity of the projectile at maximum height. $v' = v \cos 45^{\circ} = \frac{v}{\sqrt{2}}$ Angular momentum of the projectile about the point of projection $= mv'h$ $= m \frac{v}{\sqrt{2}} h = \frac{mvh}{\sqrt{2}}$

A particle is projected with a speed v at 45 ° with the horizontal. The magnitude of angular momentum of the…

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