$A$ body of mass $m$ is thrown with velocity $u$ from the origin of a coordinate system at an angle $\theta$ with the horizontal. The magnitude of the angular momentum of the particle about the origin at the time $t$ when it is at the maximum height of the trajectory is proportional to

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:

$A$ thin rod of mass $M$ and length $L$ is struck at one end by a ball of clay of mass $m$,moving with speed $v$ as shown in the figure. The ball sticks to the rod. After the collision,the angular momentum of the clay-rod system about $A$,the midpoint of the rod,is:

$A$ particle is moving in a circular path of radius $a$ with a constant speed $v$ as shown in the figure. The centre of the circle is marked by $C$. The angular momentum about the origin $O$ can be written as:

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

The position vectors of two points are $2\hat i + \hat j + \hat k$ and $2\hat i - 3\hat j + \hat k$,and the linear momentum is $2\hat i + 3\hat j - \hat k$. The angular momentum is: