Suppose that you want to launch an object that travels the longest distance, given the starting speed (or the force applied). You have to determine the angle that al you to reach the longest distance, before it touches the ground.

For simplicity, the launch happens at the ground level, so no change in altitude happens (same gravitational potential energy at the start and at the end).

In the case of a flat surface and with no air resistance, I would say that the optimal angle is 45° (and it should be easy to determine). But how the earth curvature plays in the calculation?

Optional: how can you introduce the effect of air resistance, without making the problem too complex? Is it possible to model it in a simple way, such as an arbitrary force depending only on the speed, mantaining the problem solvable? It would be also sufficient to say if the angle is unaffected or goes up/down, with some justification.

For simplicity, assume that the object has to remain in the atmosphere, so putting it in the orbit is not an option :)

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    $\begingroup$ I don't think that one can obtain analytic questions for air resistance, as one gets nasty differential equations. On the other hand, one can find the analytical solution in case of Earth curvature if you solve your problem in spherical coordinate system (as you probably know, you should get elliptical curves, even if projectile falls back on the ground - the part of ellipse is simply below ground). $\endgroup$
    – Pygmalion
    May 14, 2012 at 6:28
  • $\begingroup$ @Pygmalion and considering the air resistance as an arbitrary constant force directed backward and proportional to the square of speed? $\endgroup$
    – clabacchio
    May 14, 2012 at 6:32
  • $\begingroup$ @clabacchio Then you should solve it numerically, as you can write the equations, but an analytical solution will not be obtained. $\endgroup$
    – Bernhard
    May 14, 2012 at 6:38
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    $\begingroup$ @clabacchio Problem is that not only magnitude but also the direction of air drag force is changing... It is a tough differential equations and I don't think there's analytical solution. $\endgroup$
    – Pygmalion
    May 14, 2012 at 6:51
  • $\begingroup$ @Pygmalion what I'm looking for is a simplified model that allows to evaluate it with some approximation: I don't need an extreme accuracy, just a qualitative evaluation that helps to understand the mechanism, more than the exact value. For instance: does the air resistance make the angle lower? Why? $\endgroup$
    – clabacchio
    May 14, 2012 at 7:01

1 Answer 1


In general air resistance makes the optimal angle lower than 45$^{\circ}$. However I don't know of any rigorous way to show this without sitting down at the computer. The argument usually used is that in the prescence of air resistance you want to minimise the energy lost to the air by reducing the time of flight, and this means choosing a lower trajectory. The reduction in range by using a lower angle is balanced out by the greater average speed of the projectile.

If you Google for "optimum projectile angle air resistance" or something like this you'll find lots of articles going into this question.

  • $\begingroup$ IIRC during World War I they observed that with the massive guns (Big Bertha and such) the maximum range is obtained with angles above 45 degrees. The reason being that with a very high muzzle velocity the projectile reaches high enough altitudes such that the reduction in air drag due to lower density more than compensates for the other factors. In my youth I saw something like this in a crude simulation. But my model for the air drag (proportional to the density as well as the squared velocity, the density dropping exponentially) probably was not good enough. $\endgroup$ Oct 31, 2021 at 6:17

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