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 acceleration due to gravity)
Answer: (C) $\dfrac{mv_0^3}{4\sqrt2g}$ along negative $z$-axis
At the top, the velocity is horizontal, $v_x = \dfrac{v_0}{\sqrt2}\hat i$, and the height is
$$H = \frac{v_0^2\sin^2 45^\circ}{2g} = \frac{v_0^2}{4g}$$
$$\vec L = m\,\vec r\times\vec v = m(x\hat i + H\hat j)\times v_x\hat i = -mHv_x\hat k$$
$$|\vec L| = m\cdot\frac{v_0^2}{4g}\cdot\frac{v_0}{\sqrt2} = \frac{mv_0^3}{4\sqrt2g}$$
along the negative $z$-axis.
Solution by Sreeraj P, M.Sc Physics