Imagine that we are to throw an object at some angle with some initial velocity. Then, the distance this object will cover in the horizotnal direction is $$x = \frac{v_o^2}{g}\sin({2\alpha})$$ What's the approximated error the distance would decrease if we were to take into account the air resistance (assume no wind) ?

Let's say that this object is a bullet fired at about 300 m/s, 0.0042 kg

  • $\begingroup$ What's the object? A cannonball? A feather? $\endgroup$
    – Red Act
    Nov 20, 2016 at 14:04

2 Answers 2


If we suppose that the force of air resistance is describe by $\vec{F}_r=-bv$, where $b>0$ and $v$ is the velocity of the object, we obtain the following equations:



$$y=\frac{v_t}{g}\left(v_t+v_0\sin\alpha\right)(1-e^{-gt/v_t})-v_t t,$$

where $v_t=mg/b$. If you make $y=0$ and substitute and result of $t$ in the $x$ equation you find the distance of the projectile.

Note that if $t << v_t/g$ you obtain the expressions without air resistance.

  • $\begingroup$ This come from my universtiy notes. I don't know where my teacher took the equations or the notation. $\endgroup$
    – user326159
    Nov 20, 2016 at 17:30
  • $\begingroup$ Reference? Doesn't this come from the Farside website? $\endgroup$ Nov 22, 2016 at 4:20

A lot. It also changes the shape of the projectile's flight. With no air resistance the flight is a parabola. With air resistance the first half of the flight will be fairly parabolic, but the second half is squished.

How much the distance is shortened in the most simplistic analysis depends on the ballistic coefficient. So long as the projectile stays below about 85% of the speed of sound(about 340 m/s) then the ballistic coefficient is reasonably independent of velocity.


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