Max Throw

Suppose that the particle is projected from a height $H$ above the ground at speed $V$ at an angle $\alpha$ to the $x$-axis, with $x$ measuring the horizontal distance travelled and $y$ measuring the vertical distance travelled.
 
Then the coordinates of the points along this trajectory are
 
$$(x, y) = \left(V \cos(\alpha) t, -0.5 g t^2+V\sin(\alpha) t +H\right)\;.$$
 
The particle intersects the $x$-axis when the $y$-coordinate is zero. This is when
$$t = \frac{V\sin(\alpha) \pm \sqrt{V^2\sin^2(\alpha) +2gH}}{g}\;.$$
 
The particular case $H=0$
The square root simplifies giving the point of intersection as
$$(x, y) = \left(\frac{2V^2}{g}\cos(\alpha)\sin(\alpha), 0\right) = \left(\frac{2V^2}{g}\sin(2\alpha), 0\right)\,,$$  
where the second equality makes use of a trig identity. The $x$-value $V\sin(2\alpha)$ is maximised when $2\alpha=90^\circ$. Thus the optimal angle of projection is $45^\circ$, for any initial velocity.
 
 
When $H\neq 0$
 
In this case, the particle intersects the $x$-axis at
$$x =\frac{V^2}{g} \cos(\alpha) \left(\sin(\alpha) + \sqrt{\sin^2(\alpha) +\frac{2gH}{V^2}}\right)\;.$$
 
This looks complicated to differentiate so I tried a numerical solution using a spreadsheet.This produced an optimal angle of $40.4^\circ$ (3sf).
 
It seems clear that this angle will be dependent on the initial speed, but to check I calculate the optimum angle numerically for a large initial speed of $100\mathrm{ms}^{-1}$. This produces an optimium of $44.9^\circ$ (3sf) which is close to $45^\circ$, as we might intuitively expect.
 
Extension:

You might like to investigate this further. Here are some starting points.

Note that the expression for the derivative is:
  $$\frac{g}{V^2}\frac{dx}{d\alpha} = \cos(2\alpha)-\sin(\alpha)\sqrt{\sin^2(\alpha) +\frac{2gH}{V^2}}+\cos^2\alpha\sin\alpha \frac{1}{\sqrt{\sin^2(\alpha) +\frac{2gH}{V^2}}}\;.$$
For an optimum we can set the left hand side to zero. This gives us
$$\sin\alpha\left(\cos(2\alpha) -\frac{2gH}{V}\right) + \sin(\alpha)\cos(2\alpha)\left(1 +\frac{2gH}{V^2\sin^2(\alpha)}\right)^{\frac{1}{2}}=0\;.$$


If $X(\alpha) \equiv \frac{2gH}{V^2\sin^2(\alpha)}$ is small then we can expand to give
  $$\sin\alpha\left(\cos(2\alpha) -\frac{2gH}{V}\right)+ \sin(\alpha)\cos(2\alpha)\left(1+\frac{1}{2}\frac{2gH}{V^2\sin^2(\alpha)}+\mathcal{O}(X^2)\right)=0\;.$$
In principle that can now be turned into a polynomial expression in $\cos(\alpha)\ldots$