The trajectory of a projectile in a vertical plane is given by $y=ax-b{x}^2,$ where $a$ and $b$ are constants and $x$ and $y$ are respectively horizontal and vertical distances of the projectile from the point of projection. The maximum height attained by the particle and the angle of projection from the horizontal are

The correct option is C $\frac{a^2}{4b},{\tan }^{-1}\left(a\right)$
Trajectory of projectile in a vertical plane is given as
$y=ax-b{x}^{2}y=ax(1-\frac{x}{\left(\frac{a}{b}\right)})....\left(i\right)$
Compare $\left(i\right)$ with $y=x\tan \theta [1-\frac{x}{R}]$
$\Rightarrow \tan \theta =a$ and $R=\frac{a}{b}$
$\therefore \theta ={\tan }^{-1}\left(a\right)$

Now, For maximum height $H,$ we have
maximum height, $H=\frac{u^2{\sin }^2\theta }{2g}$ and Range $R=\frac{u^2\sin 2\theta }{g}$
So, we have the ratio $\frac{H}{R}=\frac{\frac{u^2{\sin }^2\theta }{2g}}{\frac{u^2\sin 2\theta }{g}}$
$\Rightarrow R\tan \theta =4H$
$\Rightarrow \frac{a}{b}\times a=4H\Rightarrow H=\frac{a^2}{4b}$