A particle of mass 'm' is projected with velocity 'v' making

Question

A particle of mass 'm' is projected with 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

Options:

A . Zero

B . mv34g

C . mv3√2g

D . m√2gh3

Answer: Option D
:
D

L = m v_{x} h

\Rightarrow L = m (v cos 45^{\circ}) \frac{v^{2} {sin}^{2} 45^{\circ}}{2 g}

\Rightarrow L = \frac{m v^{3}}{4 \sqrt{2} g}

Further

L = m v_{x} h

\Rightarrow L = m (v cos 45^{\circ}) h o

But

h = \frac{v^{2} {sin}^{2} 45^{\circ}}{2 g}

\Rightarrow v = 2 \sqrt{g h}

\Rightarrow L = \frac{m v}{\sqrt{2}} 2 \sqrt{g h} h

\Rightarrow L = m \sqrt{2 g h^{3}} .

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