Q. 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 particle about the point of projection when the particle is at its maximum height, is
A) $m\sqrt{2gh^{3}}$
B) $\frac{{mv}^{3}}{\sqrt{2}g}$
C) $\frac{{mv}^{3}}{4\sqrt{2}g}$
D) Zero
Explanation
(c) The speed of projectile is $v$ and angle of projection is$45{^\circ}$.
$v_{x} = v\cos{45{^\circ}} = \frac{v}{\sqrt{2}}$
$v_{y} = v\sin{45{^\circ}} - gt = \frac{v}{\sqrt{2}} - gt$
$At\ highest\ point\ v_{y} = 0,\ v_{x} = \frac{v}{\sqrt{2}}$
Maximum height achieved,
$H = \frac{v^{2}\ \sin^{2}45{^\circ}}{2g} = \frac{v^{2}}{4g}$
Now, angular momentum about O
$= \frac{mv}{\sqrt{2}} \bullet \frac{v^{2}}{4g} = \frac{{mv}^{3}}{4\sqrt{2g}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.