$A$ sportsman runs around a circular track of radius $r$ such that he traverses the path $ABAB$. The distance travelled and displacement,respectively,are

$A$ particle at a distance of $1 \ m$ from the origin starts moving,such that $dr/d\theta = r$,where $r$ and $\theta$ are polar coordinates. Then,the angle between the resultant velocity and the tangential velocity is

The ratio between the maximum range and the square of the time of flight in projectile motion is:

Two bodies of masses $m$ and $3m$ are rotating in horizontal circles of radii $r$ and $\frac{r}{3}$ respectively. The tangential speed of the body of mass $m$ is $n$ times that of the heavier body. If the centripetal force is the same for both,the value of $n$ is:

Range of a bullet fired at $45^o$ to the horizontal is $980 \, m$. If the bullet is fired at the same angle from a car travelling horizontally at $18 \, km/h$ towards the target,then the range will be increased by:

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.