$A$ person sitting in an open car moving at a constant velocity throws a ball vertically up into the air. The ball falls:

$A$ particle moving in a circle of radius $R$ with a uniform speed takes a time $T$ to complete one revolution. If this particle were projected with the same speed at an angle $\theta$ to the horizontal,the maximum height attained by it equals $4R$. The angle of projection,$\theta$,is then given 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.

If at any point on the path of a projectile its velocity is $u$ at an inclination $\alpha$ to the horizontal,then after what time will it move at right angles to its former direction?

The figure shows the $(x, t)$ and $(y, t)$ diagrams of a particle moving in $2$ dimensions. If the particle has a mass of $500 \, g$,find the force (magnitude and direction) acting on the particle.

$A$ particle moves with a constant speed of $5 \, ms^{-1}$ as shown in the figure. What is the change in velocity during half a revolution in $ms^{-1}$?