A

Step 1: Analyze the motion: The motion of the particle can be treated in two parts: horizontal and vertical.
The initial velocity $u$ can be resolved into horizontal ($u_x$) and vertical ($u_y$) components using:
$u_x = u \cos(\theta)$ and $u_y = u \sin(\theta)$, where $\theta$ is the angle of projection.

Step 2: Time of flight calculation:
The total time $T$ of flight until the particle hits the ground can be found using the vertical motion equation.
The vertical displacement is given by:
$h = u_y T - \frac{1}{2} g T^2$
Substituting $u_y = u \sin(\theta)$ leads to:
$h = u \sin(\theta) T - \frac{1}{2} g T^2$.
Rearranging gives a quadratic in terms of $T$:
$\frac{1}{2} g T^2 - u \sin(\theta) T + h = 0$.
Using the quadratic formula:
$T = \frac{u \sin(\theta) \pm \sqrt{(u \sin(\theta))^2 - 2gh}}{g}$.
Select the positive root, as time cannot be negative.

Step 3: Range calculation:
The horizontal range $R$ can be calculated using:
$R = u_x T = u \cos(\theta) T$.
Substitute the value of $T$ from the previous step to get:
$R = u \cos(\theta) \left( \frac{u \sin(\theta) + \sqrt{(u \sin(\theta))^2 - 2gh}}{g} \right)$.

Step 4: Simplification:
This expression can be further simplified to express $R$ in terms of the initial velocity $u$, the projection angle $\theta$, and the height $h$. In an ideal case where you can optimize the angle, this leads to the maximum range for a given height.

Conclusion: Thus, maximizing $R$ involves determining the best angle of projection. For maximum range, the angle at which maximum path is achieved given a height will equate to the optimal projection angle derived from physics equations.
Therefore, the maximum range formula involves integrating these components.