A ball is projected from ground at an angle $45^{\circ}$ with horizontal from distance $d_1$ from the foot of a pole and just after touching the top of pole it the falls on ground at distance $d_2$ from pole on other side, the height of pole is ...........

A projectile is thrown into space so as to have a maximum possible horizontal range of $400$ metres. Taking the point of projection as the origin, the co-ordinates of the point where the velocity of the projectile is minimum are

A particle covers $50\, m$ distance when projected with an initial speed. On the same surface it will cover a distance, when projected with double the initial speed ......... $m$

Show that for a projectile the angle between the velocity and the $x$ -axis as a function of time is given by

$\theta(t)=\tan ^{-1}\left(\frac{v_{0 y}-g t}{v_{0 x}}\right)$

Show that the projection angle $\theta_{0}$ for a projectile launched from the origin is given by

$\theta_{0}=\tan ^{-1}\left(\frac{4 h_{m}}{R}\right)$

Where the symbols have their usual meaning.

A cricketer hits a ball with a velocity $25\,\,m/s$ at ${60^o}$ above the horizontal. How far above the ground it passes over a fielder $50 m$ from the bat  ........ $m$ (assume the ball is struck very close to the ground)

A projectile thrown with a speed $v$ at an angle $\theta $ has a range $R$ on the surface of earth. For same $v$ and $\theta $, its range on the surface of moon will be