In a world without air, I understand they definitely would. However, with drag taken into account, I think they wouldn't. Since the drag force varies proportional to the square of speed (ignoring the change in coefficient of drag with speed), wouldn't the total impulse due to drag on a bullet fired from a gun from a specific height horizontal to ground be higher than the total impulse on an identical bullet dropped from the same height with no horizontal velocity?

Also, when bullet spin is taken into account, the bullet fired from a gun should resist change in orientation; therefore, it should maintain a small but non-zero angle of attack, and also have a lift force, correct?

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    $\begingroup$ I don't see how the spin or the drag would create lift. Can you spell out in more detail why you think it would? $\endgroup$ Dec 13, 2014 at 2:16
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    $\begingroup$ Also, have you seen this test? youtube.com/watch?v=D9wQVIEdKh8 $\endgroup$ Dec 13, 2014 at 2:18
  • $\begingroup$ @BrandonEnright Spin on a bullet causes resistance to change in orientation, so as the bullet follows a downward trajectory, instead of simply pointing in the direction of its velocity vector, it will point slightly upward relative to it because of the resistance to change (because of spin). It should have a non-zero angle of attack due to this, and therefore have a lift force. Am I mistaken? $\endgroup$
    – chbaker0
    Dec 13, 2014 at 2:19
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    $\begingroup$ @BrandonEnright Yeah, I have. I agree that the difference in time for the two bullets will be so tiny that it doesn't even really matter. I'm just asking if there would be a tiny difference. $\endgroup$
    – chbaker0
    Dec 13, 2014 at 2:20
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    $\begingroup$ I think there would be a (tiny) difference for a rapidly spinning bullet that is dropped versus a randomly tumbling bullet that is dropped simply because as you state, a spinning bullet resists orientation changes. I don't think this difference is enough to be meaningful in real world situations though. I seriously doubt a bullet fired from a gun would have a positive angle of attack that would allow it to convert forward motion into lift. $\endgroup$ Dec 13, 2014 at 2:29

2 Answers 2


Just based on the quadratic drag of air, yes, the fired bullet would take longer to hit the ground.

Just consider the vertical force caused by the air friction:

$F_y = - F_{\rm drag} \sin \theta = - C (v_x^2 + v_y^2) \frac{v_y}{\sqrt{v_x^2 + v_y^2}} = - C v_y \sqrt{v_x^2 + v_y^2}$

Where $\theta$ is the angle above the horizon for the bullet's velocity, and $C$ is some kind of drag coefficient. Note that when the bullet is moving down $\theta$ is negative, as is $v_y$, so the overall vertical force is positive and keeps the bullet off the ground for slightly longer.

In the dropped case, $v_x = 0$, so we get $F_y = -C v_y^2$.

In the fired case, we can neglect $v_y$ in the radical (assuming it's much smaller than $v_x$) and we get $F_y \approx -C v_y |v_x|$.

In other words, the upward force on the fired bullet is stronger, by a factor of $v_x / v_y$.

So freshman-level physics is wrong, at least according to sophomore-level physics.

Bonus Case:

If you're assuming a flat surface on earth, it's worth considering that many "flat" things (like the ocean) actually curve down and drop off below the horizon. In case you want to account for this curvature, it may be worth going to the bullet's reference frame with $\hat{y}$ always defined to point away from the center of the earth. Note that this puts you in a rotating reference frame, and then look at the centrifugal "force":

$F_y = m a = m R \omega^2 = m R \left(\frac{v_x}{R}\right)^2 = m \frac{v_x^2}{R} $

Where $R$ is the radius of the earth and $m$ is the mass of the bullet. So again, an upward force, this time proportional to $v_x$ squared. Of course this is the same as pointing out that the earth curves away from a straight line, but it's another fun application of not-quite-freshman physics.

Now you can add in much more complicated aerodynamics, but there the question sort of looses its undergrad physics charm there and becomes an aerospace engineering question!

  • $\begingroup$ Now wait a second, you're not suggesting that the "v" in the formula for drag is the magnitude of the velocity, right? Because I'm pretty sure if I fired a bullet straight downward, the drag force in the horizontal direction would be zero... $\endgroup$
    – levitopher
    Dec 13, 2014 at 3:50
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    $\begingroup$ ... or to be more precise, the horizontal force will affect the time for the bullet to hit the ground in that it modifies v_x, but the point is that the vertical force in the fired case is always greater than in the dropped case. $\endgroup$
    – Shep
    Dec 13, 2014 at 4:01
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    $\begingroup$ I do not think your first equation is correct. The drag force in the y-direction should only depend on the velocity in the y-direction, which is the same for both cases. I'll do another answer. I admit to not being an expert, perhaps we can reason it out. $\endgroup$
    – levitopher
    Dec 13, 2014 at 4:54
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    $\begingroup$ what you say is only true in the linear drag case (which dominates in, for example, most biophysics applications). In the case of quadratic drag it breaks down. $\endgroup$
    – Shep
    Dec 13, 2014 at 5:24
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    $\begingroup$ I completely agree with @fgrieu: in reality it's going to be very hard or impossible to measure anything like this effect. Just the turbulence associated with something traveling 1 km a second is liable to wash out any measurable effect. But the OP asked about quadratic drag specifically (and lift associated with the shape of the bullet, but that's going to be very complicated), not a situation where every experimental uncertainty was taken into account. But yeah, in the experimental case I completely agree, this would be very difficult to measure. $\endgroup$
    – Shep
    Dec 13, 2014 at 16:06

I don't deal with drag forces often, but I think the equation for drag is


where $F_D$ is in the same direction as $v$, and $C$ contains all the various things – density of air, cross-section, drag coefficient, etc. Importantly, $C$ depends on the orientation of the object. What I am going to do is assume the bullet falls without rotating – so it stays parallel to the ground during its entire motion (in both cases you drop it in the same direction you shoot it).

In the first case the equation of motion is found via Newton's second law:

$$\Sigma F_y=F_{D,y}-F_g=ma_y\rightarrow a_y=\frac{C_yv_y^2}{m}-g$$

In the second case, we need to consider both directions:

$$\Sigma F_y=F_{D,y}-F_g=ma_y\rightarrow a_y=\frac{C_yv_y^2}{m}-g$$ $$\Sigma F_x=-F_{D,x}=ma_x\rightarrow a_x=-\frac{C_xv_x^2}{m}$$

So to find the time of flight of either case one would have to integrate the $y$ equation, but in either case it is the same. Therefore, the time of flight for these two situations is the same. But of course, I'm assuming the bullet does not rotate during its motion.

If it did rotate, then the value of $C$ would be constant – it would be $C_x$, because that's the direction of motion – and $F_D$ would be in the direction of motion of the bullet, and $v$ would be the speed. In this case I believe the other answer would be correct, and they would reach the ground at different times.

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    $\begingroup$ It wouldn't make sense for it to be calculated by component, because then the net drag force would be calculated to be different numbers in different reference frames. It is definitely the magnitude. And, there is no contradiction; the drag force is parallel and opposite the velocity. If there is no motion along the x axis, there will be no force along the x axis. $\endgroup$
    – chbaker0
    Dec 13, 2014 at 5:09
  • $\begingroup$ Well forces being different in different reference frames is not a problem. If it's "definitely the magnitude", then I think we need to know if the bullet turns while falling. Specifically, I agree that the drag force opposite the direction of motion should be dependent on the speed - but if the bullet doesn't turn while falling, then the drag force in the y-direction will be $C_yv^2$, whereas if it turns it will be $Cv^2sin\theta$, as in the other answer. $\endgroup$
    – levitopher
    Dec 13, 2014 at 5:16
  • $\begingroup$ you can only separate the drag forces into components when you're dealing with linear drag, you're doing something wrong between your first and second equations $\endgroup$
    – Shep
    Dec 13, 2014 at 5:20
  • $\begingroup$ You're wrong in assuming you can calculate the force in the x and y directions independently. Your argument for it doesn't quite work, because the force in the horizontal direction is not $c|v|^2$. That's the magnitude of the total drag force applied to the bullet, but the direction of the force is always opposite to the bullet's direction of motion (relative to the air), so in the case of a dropped bullet, the vertical component is $-c|v|^2$, and the horizontal component is zero. $\endgroup$
    – N. Virgo
    Dec 13, 2014 at 5:22
  • $\begingroup$ The value for $C$ is different for our different cases, and that's my mistake. I'm assuming the bullet is falling like you dropped it horizontally - I'll correct it so the answer is clear in my assumptions. $\endgroup$
    – levitopher
    Dec 13, 2014 at 15:34

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