Question

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

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

Answer 1

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

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

On comparing with the general equation of trajectory,

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

we have,

$\tan \theta =2\Rightarrow \cos \theta =\frac{1}{\sqrt{5}}$

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

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

$\therefore v_0=\frac{5}{3}{\text{ms}}^{-1}$

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