Extra height for artillery is greatly prized in military tactics. As I understand it maximum range neglecting air resistance is always achieved by firing at a 45 degree angle. So what is the increased range as a function of firing height? M777 howitzers with a muzzle velocity of 827m/s can be the specific case.
And what impact will air resistance and real-world factors have?
The extra height will give you more travel time before it strikes the ground, and hence more time to travel horizontally. In an ideal situation, the projectile maintains its horizontal speed until it hits the ground. If the ground is further "down" from where it was fired, it means it will travel further, all else being equal.
As far as the optimal launch angle in this situation, it will be less than 45° as that corresponds to the ideal angle for level ground.
Mathmaticall if the launch height is $h$ above ground, then the time for impact is
$$ t = \sqrt{ \left( \frac{v \sin \theta}{g}\right)^2 + \frac{2 h}{g} } + \frac{v \sin \theta}{g} $$
and so the range is
$$ x = v \cos\theta \sqrt{ \left( \frac{v \sin \theta}{g}\right)^2 + \frac{2 h}{g} } + \frac{v^2 \sin \theta \cos \theta}{g} $$
Now the optimal angle for launch maximizes the range, and it comes out to be (in degrees)
$$ \theta° = 90° - \arctan \left( \sqrt{1+ \frac{2 g h}{v^2}} \right) $$
You can see when $h=0$, the expression above is $\theta° = 90° - \arctan(1) = 45°$. For positive values of $h$ the angle becomes less than 45°.
Assuming the projectile is fired at a 45 degree angle. Then the projectile will have an additional horizontal and vertical kinetic energies:
$$KE_h=KE_v=\frac{KE_T}{\sqrt{2}}=mgh=\frac{\sqrt{2}}{4}mv^2$$
Where $h$ is the height above the ground of the artillery. Using equations of kinematic motion, we can find that:
$$\frac{1}{2}gt^2+\sqrt{2\sqrt{2}gh}t-h=0$$
Solving for $t$, we find that:
$$t=(\sqrt{2^\frac{1}{2}+1}-2^\frac{1}{4})\sqrt{\frac{2h}{g}}$$
Which ultimately will increase the distance as:
$$d_+=v_ht=2^{\frac{5}{4}}h(\sqrt{2^\frac{1}{2}+1}-2^\frac{1}{4})\approx .86h$$
Looking at this result, we can clearly see that the horizontal distance travelled by the projectile increases proportionately to the height added to the artillery's position, which makes intuitive sense.
Hopefully this answer's your question, and hopefully my math didn't get bungled up anywhere.
