A particle of mass $m$ is projected at $45^o$ at $V_0$ speed from point $P$ at $t = 0$. The angular momnetum of particle about $P$ at $t = \frac{V_0}{g}$ is:-

A particle of mass $m$ moves in the $XY$ plane with a velocity $v$ along the straight line $AB.$ If the angular momentum of the particle with respect to origin $O$ is $L_A$ when it is at $A$ and $L_B$ when it is at $B,$ then

A particle of mass $m$ projected with a velocity ' $u$ ' making an angle of $30^{\circ}$ with the horizontal. The magnitude of angular momentum of the projectile about the point of projection when the particle is at its maximum height $\mathrm{h}$ is :

The direction of the angular velocity vector along

A particle of mass $1 kg$ is subjected to a force which depends on the position as $\vec{F}=-k(x \hat{i}+y \hat{j}) kgms ^{-2}$ with $k=1 kgs ^{-2}$. At time $t=0$, the particle's position $\vec{r}=\left(\frac{1}{\sqrt{2}} \hat{i}+\sqrt{2} \hat{j}\right) m$ and its velocity $\vec{v}=\left(-\sqrt{2} \hat{i}+\sqrt{2} \hat{j}+\frac{2}{\pi} \hat{k}\right) m s^{-1}$. Let $v_x$ and $v_y$ denote the $x$ and the $y$ components of the particle's velocity, respectively. Ignore gravity. When $z=0.5 m$, the value of $\left(x v_y-y v_x\right)$ is. . . . . $m^2 s^{-1}$

A particle is moving along a straight line parallel to $x-$ axis with constant velocity. Its angular momentum about the origin