Q. An object of mass $m$ is projected from origin in a vertical $xy$ plane at an angle $45^\circ$ with the $x$ axis with an initial velocity $v_0$. The magnitude and direction of the angular momentum of the object with respect to origin, when it reaches the maximum height, will be [$g$ is acceleration due to gravity]
A) $\frac{mv_0^3}{2\sqrt2g}$ along negative $z$-axis
B) $\frac{mv_0^3}{4\sqrt2g}$ along positive $z$-axis
C) $\frac{mv_0^3}{4\sqrt2g}$ along negative $z$-axis
D) $\frac{mv_0^3}{2\sqrt2g}$ along positive $z$-axis
Explanation
[Solution by Dr. Abubacker] At maximum height the velocity is horizontal, $v_x=\frac{v_0}{\sqrt2}$, and the height is $H=\frac{v_0^2\sin^2 45^\circ}{2g}=\frac{v_0^2}{4g}$. $|\vec L|=m v_xH=\frac{mv_0^3}{4\sqrt2\,g}$. Since $\vec r$ has positive $x,y$ and $\vec p$ points along $+x$, $\vec r\times\vec p$ points along $-z$. Answer (3).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.