The trajectory of a projectile in a vertical plane is $y=\alpha x-\beta x^2$, where $\alpha$ and $\beta$ are constants and $x$ and $y$, are respectively the horizontal and vertical distances of the projectile from the point of projection. The angle of projection $\theta$ and the maximum height attained $H$ are respectively given by

The correct option is A

${\tan }^{-1}\left(\alpha \right)$, $\frac{{\alpha }^2}{4\beta }$

Step 1: Find maxima of given equation

Given, trajectory of the projectile is described by the equation, $y=\alpha x-\beta x^2$

When maximum height ($H$) is achieved, $y$ in the above equation is also maximum because $y$ denotes the vertical position of the projectile and $x$ denotes its horizontal position.

At maximum height, $\frac{dy}{dx}=0$,
$\begin{array}{cc}\Rightarrow & \alpha -2\beta x=0\\ \Rightarrow & x=\frac{\alpha }{2\beta }\end{array}$
So the projectile reaches the highest point at $\frac{\alpha }{2\beta }$. The corresponding height will be,
$\begin{array}{rcl}y& =& \alpha \left(\frac{\alpha }{2\beta }\right)-\beta {\left(\frac{\alpha }{2\beta }\right)}^2\\ & =& \frac{{\alpha }^2}{2\beta }-\frac{{\alpha }^2}{4\beta }\\ & =& \frac{{\alpha }^2}{4\beta }\end{array}$

Therefore, maximum height $H_{max}=\frac{{\alpha }^2}{4\beta }$

Step 2: Apply formula for angle of curve at a point

The angle between a parabola and the $x$-axis at any point is given as,
$\theta ={\tan }^{-1}\left[{\left(\frac{dy}{dx}\right)}_{x=x_0}\right]$
where $\theta$ is the required angle and $x_0$ is the abscissa of the point on the parabola

At the angle of projection, $x=0$. Thus,
$\begin{array}{rcl}\theta & =& {\tan }^{-1}\left(\alpha -2\beta ·0\right)\\ \therefore \theta & =& {\tan }^{-1}\left(\alpha \right)\end{array}$

Therefore, the angle of projection, $\theta ={\tan }^{-1}\left(\alpha \right)$.

Hence, option A is correct.