When does the magnitude of angular momentum become zero?

$A$ particle of mass $m$ is projected from a point $P$ on the ground with an initial velocity $v_0$ at an angle of $45^{\circ}$ with the horizontal at $t = 0$. Find the magnitude of the angular momentum of the particle at time $t = \frac{v_0}{g}$.

An object of mass $m$ is projected from the 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 the origin,when it reaches the maximum height,will be [$g$ is the acceleration due to gravity].

If the Earth is considered a sphere of radius $R$ and mass $M$,then its angular momentum about its axis of rotation in terms of the time period $T$ will be:

$A$ particle of mass $2 \ kg$ located at the position $(\hat{i} + \hat{j}) \ m$ has a velocity $2(\hat{i} - \hat{j} + \hat{k}) \ m/s$. Its angular momentum about the $z$-axis in $kg \cdot m^2/s$ is:

$A$ particle moves along a circular path with decreasing speed. Hence,