The path of a projectile is given by $y=ax-b{x}^2$. What is the value of the angle ${\theta }_0$ and speed $v_0$ , about a point at which the projectile is launched?

The correct option is B ${\theta }_0={\cos }^{-1}\frac{1}{\sqrt{a^2+1}},v_0=\sqrt{\frac{g}{2b}(1+a^2)}$
Given that
$y=ax-b{x}^2\dots \dots \left(1\right)$

The trajectory of projectile is given by

$y=x\tan \theta (1-\frac{x}{R})$

Equation $\left(1\right)$ can be re written as

$y=ax(1-\frac{bx}{a})$
$\Rightarrow y=ax(1-\frac{x}{\frac{a}{b}})\dots \dots \left(2\right)$

Comparing equation $\left(1\right)$ with equation $\left(2\right)$ we get, $\tan \theta =a$ and Range of projectile, $R=\frac{a}{b}$

Using, $1+{\tan }^2\theta =\frac{1}{{\cos }^2\theta }$ , we can write that,

$\cos \theta =\frac{1}{\sqrt{a^2+1}}$
$\Rightarrow {\theta }_0={\cos }^{-1}\frac{1}{\sqrt{a^2+1}}$

and, we know that, $R=\frac{u^2\sin 2\theta }{g}$
Rewriting the above formula we get,

$R=\frac{u^2\left(\frac{2\tan \theta }{1+{\tan }^2\theta }\right)}{g}$

$\Rightarrow R=\frac{a}{b}=\frac{v_0^2\left(\frac{2\times a}{1+a^2}\right)}{g}$

$\Rightarrow \frac{a}{b}=\frac{v_0^2}{g}\left(\frac{2a}{1+a^2}\right)\Rightarrow v_0^2=\frac{g}{2b}(1+a^2)$

$\Rightarrow v_0=\sqrt{\frac{g}{2b}(1+a^2)}$

Thus, option (b) is the correct answer.