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

Solution :
Comparing the given equation with the equation of trajectory of a projectile,
$$\eqalign{ & y = x\tan \theta - \frac{{g{x^2}}}{{2{u^2}{{\cos }^2}\theta }} \cr & {\text{we get,}}\,\,\tan \theta = \frac{1}{{\sqrt 3 }} \Rightarrow \theta = {30^ \circ } \cr & {\text{and}}\,\,2{u^2}{\cos ^2}\theta = 20 \Rightarrow {u^2} = \frac{{20}}{{2{{\cos }^2}\theta }} = \frac{{40}}{3} \cr & {\text{Now,}}\,\,{R_{\max }} = \frac{{{u^2}}}{{\;g}} = \frac{{40}}{{3 \times 10}} = \frac{4}{3}\;m \cr} $$