$A$ ball of mass $160 \, g$ is thrown up at an angle of $60^{\circ}$ to the horizontal at a speed of $10 \, m/s$. The angular momentum of the ball at the highest point of the trajectory with respect to the point from which the ball is thrown is nearly $\left(g=10 \, m/s^{2}\right)$ (in $kg \cdot m^{2}/s$).

Show that for a projectile,the angle between the velocity and the $x$-axis as a function of time is given by $\theta(t) = \tan^{-1}\left(\frac{v_{0y} - gt}{v_{0x}}\right)$.
Show that the projection angle $\theta_0$ for a projectile launched from the origin is given by $\theta_0 = \tan^{-1}\left(\frac{4h_m}{R}\right)$,where the symbols have their usual meanings.

$A$ particle is projected vertically upwards from $O$ with velocity $v$ and a second particle is projected at the same instant from $P$ (at a height $h$ above $O$) with velocity $v$ at an angle of projection $\theta$. The time when the distance between them is minimum is

The velocity of a particle moving in a curvilinear path in a horizontal $X-Y$ plane varies with time as $\vec{v} = (2t\hat{i} + t^2\hat{j}) \ m/s$. Here,$t$ is in seconds. At $t = 1 \ s$:

$A$ particle of mass $m$ is performing $U.C.M.$ along a circle of radius $r$. The relation between centripetal acceleration $a$ and kinetic energy $E$ is given by

If a particle of mass $m$ covers half of the horizontal circle with constant speed $v$,then the change in its kinetic energy is