Q. A particle of a mass $m$ is projected with 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 (where $g =$ acceleration due to gravity)
A) Zero
B) $mv^{3}/(4\sqrt{2}g)$
C) $mv^{3}/(\sqrt{2}g)$
D) $mv^{2}/2g$
Explanation
(b) $\overrightarrow{L} = \overrightarrow{r} \times m\overrightarrow{\upsilon} = H\ mv\cos\ \theta = \frac{\upsilon\sin^{2}\theta}{2g}m\upsilon\cos\theta = \frac{m\upsilon^{3}}{4\sqrt{2}g}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.