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Equation of trajectory of projectile is given by $y = x/\sqrt{3} - gx^{2}/20$, where x and y are in meter. The maximum range of the projectile is
Explanation
Comparing the given equation with the equation of trajectory of a projectile, $y = x \tan \theta - gx^{2}/(2u^{2}\cos^{2} \theta)$ we get, $\tan \theta = 1/\sqrt{3} \Rightarrow \theta = 30^{\circ}$ and $2u^{2}\cos^{2} \theta = 20 \Rightarrow u^{2} = 20/(2 \cos^{2} \theta) = 40/3$ Now, $R_{max} = u^{2}/g = 40/(3 \times 10) = 4/3 m$