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

Particles $A$ and $B$ are moving with constant velocities along the $x$ and $y$ axes respectively. The graph of the separation $S$ between them with time $t$ is:

$A$ projectile is fired at a speed of $100 \, m/s$ at an angle of $37^o$ above the horizontal. At the highest point,the projectile breaks into two parts of mass ratio $1:3$,the smaller coming to rest. Then the distance of the heavier part from the launching point is ........... $m$.

The second's hand of a watch has $6\, cm$ length. The speed of its tip and the magnitude of the difference in velocities of its tip at any two perpendicular positions will be respectively:

$A$ man wants to reach from $A$ to the opposite corner of the square $C$. The sides of the square are $100\, m$. $A$ central square of $50\, m \times 50\, m$ is filled with sand. Outside this square,he can walk at a speed of $1\, m/s$. In the central square,he can walk only at a speed of $v\, m/s$ $(v < 1)$. What is the smallest value of $v$ for which he can reach faster via a straight path through the sand than any path in the square outside the sand?

$A$ particle is projected with velocity $u$ so that its horizontal range is three times the maximum height attained by it. The horizontal range of the projectile is given as $\frac{n u^2}{25 g}$,where the value of $n$ is (Given $g$ is the acceleration due to gravity).