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

Question

Dr. Abubacker · Accepted Answer

(b) The angular momentum of a particle is given by
$\overrightarrow{\mathbf{L}} = \overrightarrow{\mathbf{r}} 	imes m\overrightarrow{\mathbf{v}}$
$L = mvr\sin	heta$
[image:4_image108.emf]
From figure,
$L = rm(v\cos{45{^\circ}})\sin	heta$
$= \frac{mv}{\sqrt{2}}(r\sin	heta)$
$= \frac{mvh}{\sqrt{2}}\ \left( \because\ \sin{	heta = \frac{h}{r}} ight)$

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

Explanation

Reviewed by Dr. Abubacker

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