The equation of motion of a projectile is :  y  = 12x-$\frac{3}{4}{\mathrm{x}}^2$

The horizontal component of velocity is 3 ${\mathrm{ms}}^{-1}$. Given that g = 10 ${\mathrm{ms}}^{-2}$, what is the range of the projectile? 

 

y = 12x - $\frac{3}{4}{\mathrm{x}}^2$

$\frac{\mathrm{dy}}{\mathrm{dt}} = 12\frac{\mathrm{dx}}{\mathrm{dt}}-\frac{3}{2}\mathrm{x}\frac{\mathrm{dx}}{\mathrm{dt}}$

At x = 0, $\frac{\mathrm{dy}}{\mathrm{dt}} =12\frac{\mathrm{dx}}{\mathrm{dt}}$

If $\mathrm{\theta }$ be the angle of projection, then

$\frac{{\displaystyle \frac{\mathrm{dy}}{\mathrm{dt}}}}{{\displaystyle \frac{\mathrm{dx}}{\mathrm{dt}}}} = 12 = \tan \mathrm{\theta }$

Also, if u = initial velocity, then u $\cos \mathrm{\theta }$ = 3

Hence, $\tan \mathrm{\theta } \times \cos \mathrm{\theta } = 36 \mathrm{or} \mathrm{u} \sin \mathrm{\theta } = 36$

Range R = $\frac{{\mathrm{u}}^2 \sin 2\mathrm{\theta }}{\mathrm{g}} = \frac{2{\mathrm{u}}^2\sin \mathrm{\theta } \cos \mathrm{\theta }}{\mathrm{g}}$

              = $\frac{2(\mathrm{u} \sin \mathrm{\theta })(\mathrm{u} \cos \mathrm{\theta })}{10} = \frac{2\times 36\times 3}{10} = 21.6 \mathrm{m}$