In orbital motion,the angular momentum vector is directed:

The position vector of a particle of mass $m$ moving with a constant velocity $\vec{v} = v \hat{i}$ is given by $\vec{r} = x(t) \hat{i} + b \hat{j}$,where $b$ is a constant. At an instant,$\vec{r}$ makes an angle $\theta$ with the $x$-axis. The variation of the magnitude of the angular momentum $|\vec{L}|$ of the particle about the origin with $\theta$ will be:

Write the $SI$ unit of angular momentum and its dimensional formula.

$A$ particle of mass $100\,g$ is projected at time $t = 0$ with a speed $20\,ms^{-1}$ at an angle $45^{\circ}$ to the horizontal as given in the figure. The magnitude of the angular momentum of the particle about the starting point at time $t = 2\,s$ is found to be $\sqrt{K}\,kg\,m^2/s$. The value of $K$ is $............$ (Take $g = 10\,ms^{-2}$)

$A$ mass $m$ is moving with a constant velocity $v$ along a line parallel to the $X$-axis. Its angular momentum with respect to the origin or the $Z$-axis is:

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].