The equation of trajectory of projectile is given by $y=\frac{x}{\sqrt{3}}-\frac{g{x}^2}{20},$where x and y are in metre. The maximum range of the projectile is

Comparing the given equation with the equation of trajectory of a projectile,

$y=x\tan \theta -\frac{g{x}^2}{2u^2{\cos }^2\theta }$

We get, $\tan \theta =\frac{1}{\sqrt{3}}\Rightarrow \theta ={30}^0$

And $2u^2{\cos }^2\theta =20\Rightarrow u^2=\frac{20}{2{\cos }^2\theta }$

$=\frac{10}{{\cos }^2{30}^0}=\frac{10}{{\left(\frac{\sqrt{3}}{2}\right)}^2}=\frac{40}{3}$

Now, $R_{\max }=\frac{u^2}{g}=\frac{40}{3\times 10}=\frac{4}{3}m$