From the top of a tower of height $40\,m$, a ball is projected upwards with a speed of $20\,m / s$ at an angle of elevation of $30^{\circ}$. The ratio of the total time taken by the ball to hit the ground to its time of flight (time taken to come back to the same elevation) is (take $g=10\,m / s ^2$ )

In the motion of a projectile freely under gravity, its

A ball is projected with kinetic energy $E$ at an angle of ${45^o}$ to the horizontal. At the highest point during its flight, its kinetic energy will be

Derive the formula for time taken to achieve maximum, total time of Flight and maximum height attained by a projectile.

A particle is projected from the ground with velocity $u$ at angle $\theta$ with horizontal. The horizontal range, maximum height and time of flight are $R, H$ and $T$ respectively. They are given by $R = \frac{{{u^2}\sin 2\theta }}{g}$, $H = \frac{{{u^2}{{\sin }^2}\theta }}{{2g}}$ and $T = \frac{{2u\sin \theta }}{g}$  Now keeping $u $ as fixed, $\theta$ is varied from $30^o$ to $60^o$. Then,

A ball is hit by a batsman at an angle of $37^o$ as shown in figure. The man standing at $P$ should run at what minimum velocity so that he catches the ball before it strikes the ground. Assume that height of man is negligible in comparison to maximum height of projectile.  ........ $ms^{-1}$