Question

OA particle of mass m is projected with a velocity v mak-
ing an angle 45° with the horizontal. The magnitude of
the angular momentum of the projectile about the point
of projection when the particle is at its maximum heighth
is-​

Answer 1

Answer:

(m * v^3)/4g

Explanation:

the maximum height achieved by particle is (v× sin45°)^2/2g = v^2/4g

angular momentum = m×v×r

where r is shortest perpendicular distance from line of velocity......which in this case will be max. height