Question

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 meter. The maximum range of the projectile is

• A. $\frac{8}{3}m$
• B. $\frac{4}{3}m$
• C. $\frac{3}{4}m$
• D. $\frac{3}{8}m$

Answer 1

The correct option is B $\frac{4}{3}m$

Comparing the given equation with equation of trajectory of projectile,
$y=x\tan \theta -\frac{g{x}^2}{2u^2{\cos }^2\theta }$

We get, $\tan \theta =\frac{1}{\sqrt{3}}\Rightarrow \theta ={30}^{\circ }$
and $2u^2{\cos }^2\theta =20\Rightarrow u^2=\frac{20}{2{\cos }^2\theta }$

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

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