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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

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Correct option is B

Comparing the given equation with the equation of trajectory of a projectile,y=xtanθ−gx22u2cos2θWe get, tanθ=13⇒θ=300And 2u2cos2θ=20⇒u2=202cos2θ=10cos2300=10(32)2=403Now, Rmax=u2g=403×10=43m

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The equation of trajectory of projectile is given by y=x3−gx220,where x and y are in metre. The maximum range of the projectile is

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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