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

• A. Zero
• B. $\frac{m{v}^3}{4g}$
• C. $\frac{m{v}^3}{\sqrt{2}g}$
• D. $m\sqrt{2g{h}^3}$

Answer 1

The correct option is D $m\sqrt{2g{h}^3}$

$L=m{v}_{x}h$
$\Rightarrow L=m\left(v\cos {45}^{\circ }\right)\frac{v^2{\sin }^2{45}^{\circ }}{2g}$
$\Rightarrow L=\frac{m{v}^3}{4\sqrt{2}g}$
Further $L=m{v}_{x}h$
$\Rightarrow L=m\left(v\cos {45}^{\circ }\right)ho$
But $h=\frac{v^2{\sin }^2{45}^{\circ }}{2g}$
$\Rightarrow v=2\sqrt{gh}$
$\Rightarrow L=\frac{mv}{\sqrt{2}}2\sqrt{gh}h$
$\Rightarrow L=m\sqrt{2g{h}^3}.$