$A$ particle of mass $m$ is projected with a velocity $V$ making an angle of $45^{\circ}$ with the horizontal. The magnitude of the angular momentum of the projectile about the point of projection when the particle is at its maximum height $h$ is

  • A
    zero
  • B
    $\frac{mV^2}{\sqrt{2}g}$
  • C
    $\frac{mV^2}{4\sqrt{2}g}$
  • D
    $m\sqrt{2gh^3}$

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