Why do the bullet fired horizontally and the bullet dropped hit the ground at the same time?

With their season premiere this week, the Mythbusters are testing a classic physics story, so of course I had to comment on it. The myth in question is that if you fire a bullet from a gun held horizontally, it will hit the ground at the exact same time as a bullet dropped without any horizontal motion at all.

Of course, in the mind of any physicist, this is no myth at all — the laws of physics that tell us this should happen are so well established that they’re almost beyond question. Specifically, it’s the linear independence of orthogonal vectors, which means that components of motion that are perpendicular to each other, like gravity (vertical) and constant velocity (horizontal), don’t get in each other’s way. You can split the motion of the bullet into two perpendicular components and analyze each one separately. This is, in fact, one of the first things students learn in an introductory physics class: analyzing the motion of a fallen or thrown object. The equations $x = v_{0x}t$ and $y = -\frac{1}{2}gt^2$ work for both the fallen bullet and the dropped bullet, just with $v_{0x} = 0$ in the latter case. But that velocity doesn’t make any difference at all to the vertical motion, since $v$ doesn’t appear in the equation for $y$ at all.

Unfortunately, (practically) nothing in real-world physics is really that simple. In reality, we have air resistance to deal with. Air resistance slows down a speeding bullet, and so you’d think that it would hold the fired bullet back, so that the dropped bullet would hit the ground first. But wait! The dropped bullet is also moving through the air, and it experiences air resistance as well. So does that mean that they would hit the ground at the same time after all?

The drag force on an object moving through a fluid can be calculated as

$$F_D=\frac{1}{2}\rho v^2 A C_D$$

where $\rho$ is the density of the fluid, $v$ is the object’s velocity, $A$ is its cross-sectional area, and $C_D$ is the drag coefficient, which is typically on the order of 0.1 for streamlined shapes or 1 for rounder or blockier shapes. Wikipedia reports $C_D \sim 0.295$ for some bullets so let’s go with that. The first thing to notice is that this formula depends on velocity squared — it’s non-linear. That means that when drag force is involved, the horizontal and vertical components of velocity may not be independent anymore.

Drag force always acts antiparallel to the overall velocity of a moving object. So if the bullet is moving straight horizontally, as it starts out, the drag force will also be horizontal, but as the bullet’s trajectory tilts downward, the drag force will angle upward to remain antiparallel. The horizontal and vertical components of the drag force are

$$\begin{aligned}F_{Dx} &= \frac{1}{2}\rho v^2 A C_D \frac{v_{x}}{v} \\ F_{Dy} &= \frac{1}{2}\rho v^2 A C_D \frac{v_{y}}{v}\end{aligned}$$

and including that effect modifies the equations of motion to

$$\begin{aligned}x(t) &= v_{0x} t - \frac{\rho A C_D}{4m}v(t) v_{x}(t) t^2 \\ y(t) &= -\biggl(\frac{g}{2} + \frac{\rho A C_D}{4m}v(t) v_{y}(t)\biggr) t^2\end{aligned}$$

Clearly, this is no longer a simple set of equations. You can’t solve it exactly, but you can simulate it on a computer.

I set up a simulation of the track defined by these equations, using the Open Source Physics toolkit to do the calculation and rendering. In the show, Adam and Jamie said that they found the two bullets (fired and dropped) hitting the floor within about 36 milliseconds of each other. What the simulation can tell us is whether air resistance accounts for that difference. So does it? Well, here’s the simulated result:

Graph of height vs. time

This graph is zoomed in closely on the point where both bullets would be hitting the floor (that would be the 0 on the vertical axis). The green track represents the dropped bullet and the red track represents the fired bullet. According to this output, the dropped bullet in the simulation hits the ground after falling for $\unit{432}{\milli\second}$ and the fired bullet hits the ground after failling for slightly under $\unit{441}{\milli\second}$ — a difference of only $\unit{9}{\milli\second}$, about a quarter of what the Mythbusters observed.

So, does this mean that a dropped bullet and a fired bullet actually don’t hit the ground at the same time? Actually, we really can’t tell based on the Mythbusters’ single experiment and one simple simulation. $\unit{9}{\milli\second}$ is a pretty small time compared to the total flight time of the bullets, and if we wanted to be able to confidently compare the simulation’s result to reality, we’d need to have a much more accurate model of the forces acting on the bullet in flight. And even if our simple model is reasonably accurate, there are only a few tens of milliseconds left to be accounted for by other factors. That’s not very much. Everyday devices like the gun and the drop rig used in the experiment just aren’t designed to activate with that level of precision, so it’s easy to imagine that some sort of mechanical randomness could introduce a fraction of a second’s delay one way or the other, enough to account for the discrepancy between the experiment and the simulation. It would have been great if the Mythbusters took the time to repeat their shot to try to eliminate statistical errors from the data, but the next 20 trials probably aren’t such great television.

I was listening to Joe Rogan taking to Neil DeGrasse Tyson. They were discussing snipers & the curvature of the Earth. Then Joe says "If you are holding a bullet in your hand & drop it, and fire a gun at the same time, they will both hit the ground simultaneously" (paraphrasing)

Is that true? That seems outrageous! How does that work?

Most recent answer: 10/22/2007

If you shot a bullet parallel to the Earth’s surface and dropped a bullet at the same time from the same height, would both bullets hit the ground at the same time?- David

Clear Lake High School, Texas

Well, the simplified physics textbook problem would probably want the answer "yes," since you can treat Newton’s second law and the resulting kinematics of uniformly accelerated motion separately for the horizontal and vertical components of the motion. They probably tell you to treat the air resistance as being zero and treat the Earth as flat, both of which we know not to be true. But more realistically, air resistance provides a drag force which increases with speed and points in the direction opposite to the velocity of the bullet through the air. The bullet is traveling very rapidly, mostly horizontally, but with a small downwards component. Air resistance provides a force that increases nonlinearly with speed, and so the vertical component of the air resistance force will be greater for the horizontally shot bullet than for the dropped bullet. If the bullet is shot very fast, the curvature of the Earth becomes important (say if this is an artillery shell being shot at a target many miles away). Shoot a bullet fast enough in the horizontal direction, ignore air resistance, and you can get it in orbit around the Earth. Alternatively, if there’s a hill nearby, you could get the opposite answer. Tom J.

(published on 10/22/2007)

So what is the answer to the question??Which bullet will hit the earth first?? I’ve been arguing this with a mate for years. I think the one dropped will hit the earth before the bullet that is shot.write back to (deleted)- Dombowski (age 27)

Hong Kong

Before one can answer a yes/no, black/white kind of question that you have posed you must be extremely precise in specifying all the environmental conditions involved. Do you want to neglect air resistance, air currents, the earth’s rotation, geographical position, direction of firing, and on and on and on...? The classic high school text answer is that they both fall to the earth at the same time. However if you are at the earth’s equator the answer depends on whether you are firing toward the east or toward the west. Figure that one out... you can have more arguments with your mate.LeeH

(published on 10/22/2007)

The question of a fired vs a falling bullet is an interesting thought experiment. It would seem to me that a bullet fired with significant velocity at a perfectly horizontal angle to the immediate plane of the earth (in other words perfectly perpendicular to the earth's gravitational force) would hit the ground after the other bullet, being of identical size and mass, which is dropped from the same exact height above ground. this even assumes that the test is conducted over a topological flat ground. Bullets falling to the ground, whether by dropping or firing, are set into motion on a linear path toward the gravitational center of the earth. In this example we can ignore the slight gravitational variations the fired bullet would travel since the increase of gravitational force would be too small to appreciably affect the result. The main issue here is that by firing the bullet we have applied energy to the bullet that has acted on it in contradiction to its original linear path; we have set it on a new linear path. The two will begin to average out, forming a parabola (or bullet drop) over distance, as the two forces work against each other. But, since we have applied energy to the mass in a linear direction different than that of the linear direction of the gravitational energy, we have acted on it, which as Newton so aptly pointed out, will affect its tendency to "stay in motion". The problem with the thought experiment is that of scale. If you step back and look at it from a larger scale, like that of the bullet's relationship to the entire planet (and the spherical curvature of space-time surrounding our planet's gravitational center) it makes more sense. When you fire that bullet, you've essentially set that bullet in motion against that curvature. Satellites are a good example. Ignoring the concepts of "geostationary" as that is not relevant here, this should make sense. The way that a satellite stays in orbit is by getting it up to the desired height and applying a linear momentum to the satellite perpendicular to the earth's gravitational field. In this way, the satellite will want to travel in a straight line, conserving linear momentum, but the earth's gravity will "tether" the satellite as it exerts energy to pull the satellite along its linear path. The result is the two forces fighting each other: The satellite can't fly off into deep space because of the gravitational "tether", and the satellite won't just fall straight to the earth due to its desire to travel in a linear path perpendicular to this other force. If you balance these forces, the satellite maintains orbit. If, however, you launch the same satellite up to the same height and fail to give it any linear momentum contradicting the gravitational attraction of the two bodies, the satellite will simply fall back to earth, following its linear path along the curvature of space-time towards the gravitational center. The bullets are the same way, only on a much smaller scale. Theoretically, if you fired the bullet with sufficient energy as to have equal energy in both perpendicular paths, take out obstacles and drag coefficient, and could theoretically maintain the linear energy by adding additional energy to the fired bullet as it travels, it would essentially orbit the earth at 3 feet from the ground. It's a matter of energy, of which mass is only a portion of the equation. Velocity is the other portion of that equation. I can imagine that if the test were conducted with the use of a 50 caliber bullet or a large rail gun then the time differences would be such that it would be easier to see the differences between a dropping a bullet with no energy or momentum in the perpendicular linear path or dropping a bullet with vast quantities of energy and high velocities in the perpendicular linear path. I'm by no means a physics expert or professor so if my terminology is off then I apologize, but the concept is evident. Perhaps this angle (pun partially intended) is the one that has been overlooked in the thought experiment because it's hard to imagine a tiny bullet in scale to the entire earth and the space-time in which the earth sits and is traveling.- Michael (age 38)

Greer SC

Hello Michael,

You are right; a bullet fired horizontally will take a longer time to fall to earh. If fired with enough velocity it can orbit the earth of even go off into space. Your problem was described by Isaac Newton in the early 18th century. It goes by the name "Newton's Cannon Ball". See: https://en.wikipedia.org/wiki/Newton's_cannonball for a full explanation.

LeeH

(published on 12/30/2015)

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