The equation of motion of a projectile is:   $y = 12x - \frac{5}{9}{x^2}$. The horizontal component of velocity is $3\ ms^{- 1}$ . Given that $g = 10\ ms^{- 2}$ , .......... $m$ is the range of the projectile .

A projectile is projected at $30^{\circ}$ from horizontal with initial velocity $40\,ms ^{-1}$. The velocity of the projectile at $t =2\,s$ from the start will be $........$ (Given $g =10\,m / s ^2$ )

The range of a projectile for a given initial velocity is maximum when the angle of projection is ${45^o}$. The range will be minimum, if the angle of projection is ......... $^o$

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 particle is projected at angle $\theta$ with horizontal from ground. The slop $(m)$ of the trajectory of the particle varies with time $(t)$ as ...........

At $t = 0$ a projectile is fired from a point $O$(taken as origin) on the ground with a speed of $50\,\, m/s$ at an angle of $53^o$ with the horizontal. It just passes two points $A \& B$ each at height $75 \,\,m$ above horizontal as shown. The horizontal separation between the points $A$ and $B$ is ........ $m$