Categories > Motion in A Plane > oblique-projectile-motion > A particle of mass $m$ is projected with velocity $v$ making an angle of ${45^o}$with the horizontal. The magnitude of the angular momentum of the particle about the point of projection when the particle is at its maximum height is (where $g = $ acceleration due to gravity)

Motion in A Plane

Question: A particle of mass $m$ is projected with velocity $v$ making an angle of ${45^o}$with the horizontal. The magnitude of the angular momentum of the particle about the point of projection when the particle is at its maximum height is (where $g = $ acceleration due to gravity)


1) Zero
2) $m{v^3}/(4\sqrt 2 g)$
3) $m{v^3}/(\sqrt 2 g)$
4) $m{v^2}/2g$
Solution: Explanation: Explanation
oblique-projectile-motion

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