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?

  • $\begingroup$ Insofar as this is physics, try searching for 'projectile motion with air resistance' here or in a web browser, once you've acquainted yourself with how to solve the problem in a vacuum. Real objects don't necessarily have a single drag coefficient (for instance, projectiles may start to tumble), so empirical tables will be more useful than first principles for a specific problem like a particular gun firing a particular round. $\endgroup$
    – g s
    Commented Jun 7, 2022 at 22:09
  • $\begingroup$ The answer is in the Wikipedia article Range of a projectile and the links therein. $\endgroup$
    – Farcher
    Commented Jun 8, 2022 at 8:11
  • $\begingroup$ I would be careful about the assumption that the advantage of the higher ground is purely range advantage, and not other military factors. High ground has very distinct set of advantages, and disadvantages, that must be taken into consideration in any military deployment. $\endgroup$
    – Max R
    Commented Aug 31, 2022 at 0:44
  • 1
    $\begingroup$ A former soldier mentioned that it would be an advantage to take the high ground around Bakhmut. The range that is needed here is so short that I assume the advantage is not related to range at all. A howitzer with simple munitions has a range of 25 km, the hills in question are something like 5 km away. Without drone reconnaissance, visibility would make a big difference, and a flat trajectory could reduce the travel time (relevant because the projectiles are subsonic, so there is a warning time long enough to jump into a trench). So I think there are aspects that do not depend on physics. $\endgroup$ Commented May 30, 2023 at 21:54
  • $\begingroup$ @VolkerSiegel: Defensible heights help for directing artillery, but it's invariably not fired by sight. Kinetic energy should be higher. High ground above low ground reduces ability to hide, eg behind ridges.. Swedish Archer motorised artillery, can launch 3 sequential shots that land at the same time, by going from steep to shallow trajectory - indicating maximal range for given energy are not always used.. I think I need to ask military folks... $\endgroup$
    – CriglCragl
    Commented May 30, 2023 at 22:04

2 Answers 2


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:


Where $h$ is the height above the ground of the artillery. Using equations of kinematic motion, we can find that:


Solving for $t$, we find that:


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.

  • 1
    $\begingroup$ You should add how the maximum angle for longest range is achieved by firing from and angle smaller than 45 degrees. $\endgroup$
    – Triatticus
    Commented Jun 7, 2022 at 20:59
  • $\begingroup$ This is false. The main cause of the error is the second sentence, in which you've made lifting a cannon increase its muzzle velocity. $\endgroup$
    – g s
    Commented Jun 7, 2022 at 21:58
  • $\begingroup$ An object does not have separate horizontal or vertical kinetic energy. KE is not a vector quantity. You cannot separate the velocity components, just won't work. $\endgroup$ Commented Jun 7, 2022 at 22:09
  • $\begingroup$ @JohnAlexiou $\frac 12 mv^2 = \frac 12 m \dot x^2 + \frac 12 m \dot y^2$ $\endgroup$
    – Farcher
    Commented Jun 8, 2022 at 8:24
  • $\begingroup$ @Farcher - yet this "split" has no meaning and only exists because mass is a scalar (velocity and momentum are parallel). You cannot assert any conservation laws to any part of KE the way it is split up. $\endgroup$ Commented Jun 8, 2022 at 13:45

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