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 bullet is fired from a gun at the speed of $280\,ms ^{-1}$ in the direction $30^{\circ}$ above the horizontal. The maximum height attained by the bullet is $……..\,m$ $\left(g=9.8\,ms ^{-2}, \sin 30^{\circ}=0.5\right):-$

A projectile is fired at $30^{\circ}$ to the horizontal, The vertical component of its velocity is $80 \;ms ^{-1}$, Its time flight is $T$. What will be the velocity of projectile at $t =\frac{ T }{2}$?

The greatest height to which a man can throw a stone is $h$. The greatest distance to which he can throw it, will be

Define projectile particle and derive the equation $y\, = \,(\tan \,{\theta _0})x\, – \,\frac{g}{{(2\,\cos \,{\theta _0})}}{x^2}$