Question

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 given by which of the following relation(s)?

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

Answer 1

Answer: (B), (D)

The correct options are
B $\frac{m{v}^3}{4\sqrt{2}g}$
D $m\sqrt{2g{h}^3}$

Angular momentum($L$) = $\overrightarrow{r}\times \overrightarrow{p}$
$L$ = $mvrsin\theta$
$L$= (momentum)$\times$(perpendicular distance)


Angular momentum about origin

$L=m\frac{v}{\sqrt{2}}h$ ----(i)
$h=$ max. height of a projectile
$h=\frac{v^2{\sin }^2\theta }{2g}$
$h=\frac{v^2}{4g}$ {$\because \theta ={45}^{\circ }$} ----(ii)

from (i) $\&$ (ii),

$L=\frac{m}{\sqrt{2}}\left(2\sqrt{gh}h\right)=m\sqrt{2g{h}^3}$

$L=\frac{mv}{\sqrt{2}}\times \frac{v^2}{4g}=\frac{m{v}^3}{4\sqrt{2}g}$

Option B and D are correct.