Question

The trajectory of the projectile near the surface of the earth is given as $y=2x-9x^2.$ If it was launched at an angle ${\theta }_0$ with speed $v_0$ then $(g=10{\text{m/s}}^2)$

• A. ${\theta }_0={\sin }^{-1}\frac{1}{\sqrt{5}}\text{and}{v}_0=\frac{3}{5}{\text{ms}}^{-1}$
• B. ${\theta }_0={\cos }^{-1}\frac{2}{\sqrt{5}}\text{and}{v}_0=\frac{3}{5}{\text{ms}}^{-1}$
• C. ${\theta }_0={\cos }^{-1}\frac{1}{\sqrt{5}}\text{and}{v}_0=\frac{5}{3}{\text{ms}}^{-1}$
• D. ${\theta }_0={\sin }^{-1}\frac{2}{\sqrt{5}}\text{and}{v}_0=\frac{3}{5}{\text{ms}}^{-1}$

Answer 1

The correct option is C ${\theta }_0={\cos }^{-1}\frac{1}{\sqrt{5}}\text{and}{v}_0=\frac{5}{3}{\text{ms}}^{-1}$

Given, $y=2x-9x^2$

On comparing with,

$y=x\tan \theta -\frac{g{x}^2}{2v_0^2{\cos }^2\theta }$

We have,
$\tan \theta =2\text{or}\cos \theta =\frac{1}{\sqrt{5}}[\because \cos \theta =\frac{1}{\sqrt{1+{\tan }^2\theta }}]$

$\text{And,}\frac{g}{2v_0^2{\cos }^2\theta }=9$

$\Rightarrow \frac{10}{2v_0^2(1/\sqrt{5}{)}^2}=9$

$\therefore v_0=\frac{5}{3}\text{m/s}$

Hence, $\left(C\right)$ is the correct answer.