Question

The equation of trajectory of a projectile is $y=(\frac{x}{\sqrt{3}}-\frac{g{x}^2}{20})$, where $x$ and $y$ are in meter. The maximum range of the projectile is (take $g=10m/s^2$)

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

Answer 1

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

The given equation of trajectory of projectile is $y=(\frac{x}{\sqrt{3}}-\frac{g{x}^2}{20})$........(i)
We know that the equation of trajectory is
$y=x\tan \theta -\frac{g{x}^2}{2u^2{\cos }^2\theta }$...........(ii)

Comparing eq (i) & (ii) we get
$\tan \theta =\frac{1}{\sqrt{3}}\Rightarrow \theta ={30}^{\circ }$
Also, $2u^2{\cos }^2\theta =20$
$\Rightarrow u^2=\frac{20}{2{\cos }^2{30}^{\circ }}=\frac{40}{3}{m}^2/s^2$
As we know that the maximum range of the projectile is given by
$R_{max}=\frac{u^2}{g}=\frac{40/3}{10}=\frac{4}{3}\text{m}$