Question

The equation of a projectile is given by $y=\sqrt{3}x-5x^2$. The range of the projectile is

• A. $\frac{\sqrt{3}}{5}\text{m}$
• B. $\frac{2\sqrt{3}}{3}\text{m}$
• C. $\frac{3\sqrt{2}}{2}\text{m}$
• D. $\frac{\sqrt{5}}{3}\text{m}$

Answer 1

The correct option is A $\frac{\sqrt{3}}{5}\text{m}$

We know the general equation of motion of projectile:
$y=x\tan \theta -\frac{g{x}^2}{2u^2{\cos }^2\theta }$

$y=x\tan \theta (1-\frac{gx}{2u^2{\cos }^2\theta .\tan \theta })$

$y=x\tan \theta (1-\frac{gx}{2u^2\cos \theta .\sin \theta })$

$y=x\tan \theta (1-\frac{gx}{u^2\sin 2\theta })$

$y=x\tan \theta (1-\frac{x}{R})....\left(1\right)$,
as $R=\frac{u^2\sin 2\theta }{g}$

we got,
$y=\sqrt{3}x-5x^2$
$y=\sqrt{3}(x-\frac{5x^2}{\sqrt{3}})......\left(2\right)$
Comparing $\left(1\right)$ and $\left(2\right)$ we get
$R=\frac{\sqrt{3}}{5}m$