Rhett Allain is an Associate Professor of Physics at Southeastern Louisiana University. He
enjoys teaching and talking about physics. Sometimes he takes things apart and can't
put them back together.If you didn't catch the latest MythBusters (yeah! new episodes), they did something straight from the physics textbooks. Just about every text has this example of shooting a bullet horizontally and dropping a bullet from the same height. The idea is that they should hit the ground at the same time. No one but the MythBusters could actually show this demo with a real gun.
The Physics
I am going to do some calculations, but I want to first write about the physics that accompanies this idea (and you can actually do it your self without the gun). What physics principle does this demo show? Well, it shows two things. First, it shows that the horizontal and vertical motion in projectile motion are independent. That is, what happens in the y-direction, stays in the y-direction. Really, you can treat projectile motion as two separate problems that just happen to take the same amount of time. The second thing it shows is that the velocity in one direction does not effect the velocity in the other direction (which really is the same as the first thing). Let me draw two diagrams, one for a dropped ball and one for ball shot horizontally.
There are some really important things to note in this diagram. First, the force is only in the y-direction. This means that the red ball's x-motion does not change and neither does the blue ball's. Second, the acceleration in the y-direction for both balls will be the same. Even if the balls had different masses, this would be true since a greater mass would have a greater gravitational force from the Earth.
Since the x- and y-motions are independent, and since the balls have the same y-motion they would hit the ground at the same time.
But what about air resistance?
Yes, this complicates things somewhat. First, I will use a typical model for air resistance that depends on some stuff and the speed of the object. Here is the model:
This is the magnitude of the air resistance. The direction is in the opposite direction as the way the object moves. So, here is a bullet shot horizontally from a gun with air resistance.
Here there is a difference. Now there is a component of force that depends on velocity in the y-direction. Will this make it hit the ground at a different time than one that is just dropped? Even if that one also has air resistance? Let's just do it. Of course, I am not the MythBusters. I don't even own a gun. I do own a computer.
Modeling Bullet with air resistance
It seems like I have been using my bullet model a lot lately. Each time I change it just a little. In this case, I am going to start with a .45 caliber bullet like the mythbusters did. They said they used it because it was heavy and slow. That is fine with me. Slow means that I can just use one value for the drag coefficient instead of my super-sonic coefficient table. If you want my python program, I will add it at the end. But here is my first run. I am going to shoot a .45 caliber bullet aimed horizontally 36 inches above the ground. Just as a comparison, I will plot the trajectory of the .45 and another object shot with it but without air resistance.
This gives a range of 102 meters (or feet) which is similar to what the MythBusters found. Notice that without air resistance, the object only goes a few (7 ish) meters more. Now how about a comparison of their vertical motions.
I zoomed in to just the end. Here you can see that the bullet with air resistance hit the ground after 0.44 seconds. The bullet without air resistance hit the ground after 0.431 seconds. Ok, now to add air resistance to the dropped bullet and just drop it (don't shoot it). Also, I will add a plot of a falling bullet without air resistance.
They don't hit at the same time. Actually, all three objects will hit differently. The no air drag object will hit first, then the dropped bullet and then the fired bullet. The above graph looks like just two lines because the dropped bullet and the no drag bullet are so similar. Let me zoom in a bit on that graph.
What if I increase the air resistance? I could redo this with something like a baseball, but I will just increase the coefficient C. Here is what I get.
This is with a ginourmous drag coefficient of C = 20. So, is this right? Should the dropped bullet really hit the ground first with air resistance? Let me draw two force diagrams for a dropped bullet and a fired bullet just a short moment after they are released.
When initially released, both bullets had the same force in the y-direction. The fired bullet was moving horizontally so that the air resistance was only in the x-direction. So, they both initially had the same vertical acceleration and after a short time about the same velocity (I will call this v2y). But, how do the vertical air resistances compare? Let me call the initial launch speed of the bullet vbullet. One more thing, I am going to write the air resistance as Kv2 just because it is shorter. The air resistance on the dropped bullet will be:
Now let me look at the y-component of air resistance force on the fired bullet. Let me just say that it really hasn't slowed down much (doesn't really matter anyway) in the x-direction. The magnitude of the whole air resistance force would be:
I left off that tiny little y-component of velocity because it is just too small to matter. But what I want is the vertical component of the air resistance. To do this, I need the angle below the horizontal the bullet is traveling. This diagram may be useful:
So, I can write the following:
And then the y-component of the air resistance would be:
There you have it. The y-component of air resistance for the fired bullet still depends on the fired speed of the bullet (since it is proportional to v2). A fired bullet (with air resistance) does not hit the ground at the same time as a dropped bullet. That being said, it was still an awesome demo. The .45 caliber was a great choice since it is slow and heavy and air resistance doesn't play that big of a role.
Update
I forgot to post my sloppy python code for this problem. Here it is:
Comments
If you fire horizontally the geoid curves downward L^2/8R to first order, "L" being pathlength and "R" being the planet's radius (about 3961 miles or 20,924,000 feet). Over 335 feet of flight the bullet had an added drop of about 7 mils (presumably taxable).
Posted by: Uncle Al | October 10, 2009 3:50 PM
Fantastic post (and blog) you've got here :-) I actually found this in the course of writing up nearly the same thing for my own blog. I was wondering if you'd consider sharing the code you used to develop those graphs? I'm running my own simulations and consistently finding a drag force which is much larger than it should be - I get the bullet flying only about 25m (using rho=1.2, A=0.0001, m=0.015, C=0.2, all in MKS units), and It'd be great to be able to check my program against yours.
Posted by: David | October 11, 2009 6:44 PM
@David,
I updated the page and included my python code. I am completely open to the idea that there is a problem with my code (it happens).
Posted by: Rhett | October 11, 2009 7:33 PM
One thing that has always perturbed my in these physics demos is that the don't reflect reality at all. Bullet arcs form parabolas as described above, but they don't start at the peak of a parabola. The bullet arcs up from the point of the barrel before arcing down again.
You can prove this by keeping the gun steady, firing at a paper target, move the target forward, then fire again. The second target will be higher by a proportion of the distance you moved forward. That proportion also depends on the calibre of the weapon fired.
I once asked my physics professor why this was so back when I was in college. He wouldn't believe me. He asked around, and apparently it has something to do with fluid dynamics. But it's simple enough to prove but most physics people won't believe you. So, one other reason a fired bullet won't hit the same time a dropped bullet will. It has vertical, upword movement a dropped bullet does not.
Posted by: Nick | October 12, 2009 11:05 AM
@ Rhett
No, it was a problem with my code, - I had an extra factor of v. Thanks for adding your program!
Posted by: David | October 13, 2009 12:52 AM
you forgot about the riffling found on modern guns which will increase the airtime of bullets. The arc of the earth, rotational inertia, and the aerodynamics on the bullet itself are too many variables for the myth to hold true. Also remember that warm air (and warm air around a hot bullet will rise when surrounded by colder air) which may account for some of the lift found on modern weapons; The spinning bullet will have colder (and denser air) below it, but warmer(less dense air) above it.
Posted by: Tyler | October 14, 2009 7:46 PM
I enjoyed the analysis of the physics on this site but the bottom line is that the fired bullet hit the ground 9.2% later than the dropped bullet… a significant amount! (.0396 sec/.4312 sec)=.092. myth busted
Posted by: calvin | October 15, 2009 1:01 PM
The problem is always presented in text books with the proviso, "negligible air resistance".
Posted by: Mark | October 16, 2009 9:34 PM
Not that any additional confirmation is needed but I set the problem up in Mathematica using a Runge Kutta instead of the Euler algorithm and got essentially the same result. There is a breath over 2% difference in the landing times in the calculations. I didn't see the original video of the Mythbusters but from the youTube short I gather they saw a wider difference. Some of that may simply have been timing error in their equipment (trying to pick out a speck on a high speed camera). I also note that the pistol had a significant kick as the bullet emerges. This could have given the bullet a slight upward velocity. I think this one is close enough to the textbooks to call it good.
Posted by: paul | October 19, 2009 11:19 PM
The fun with this, that actually the measurement in the show did prove the myth to be false, but they said, oh, 40 ms difference is just so small, so let it be confirmed.
Posted by: Bert | October 22, 2009 4:35 AM
To Paul I reply I agree that the methodology may have been flawed. The gun did kick causing an upward flight, the dropping of the bullet may have been mistimed, and the evaluation of when the bullets hit the ground may have been off. To Bert I reply yes that the difference was only 40 ms at 360 feet, but what would it have been at a quarter mile, the distance many bullets would travel. If the difference were 200 ms would that change your decision? It’s not the time that counts but the percentage. I suspect that the fired bullet really does “fly”! I base this idea on the experience of flying model airplanes. An airplane with a fully symmetrical airfoil (the top contour of the wing is identical to the bottom contour) can fly upside down. A bullet is similar to a symmetrical wing. I believe a fired bullet does have some lift keeping it in the air for a longer period of time. I would like to see this test repeated with better methodology. Which would determine conformation or busting.
Posted by: calvin | October 22, 2009 4:50 PM
@Nick
That's because in your examples you haven't been aiming exactly horizontally. You cannot rely on the sights alone to determine the direction the barrel is pointing, as sights are generally compensated for drop and parallax for a determined distance. A rifle calibrated to hit a 50 meter target will not hit the point of aim at most other distances. The further you calibrate sights, the greater the angle between the horizontal and the barrel. To see your initial trajectory, you must use a laser bore sight. No crazy fluid dynamics.
Posted by: Warren | October 26, 2009 1:03 AM
Read your article with interest. Would be interested in opinions of the rising fastball myth. Perhaps you saw the episode on Mythbuster's too. It appears that you guys may have some additional insight to provide in favor of the rising fastball.
Take 100 mph baseball traveling 60 feet from mound to plate from a height of 6 feet. Previous models include gravity and Mangus effect and current model include limited drag, but no discussion of lift or air resistance. IMO, it has less to do with the Bernoulli principal of an 76mph curveball and more to do with the spinless flutter of a 76mph knuckleball. Please help me with the physics.
From a statistical standpoint, I have determined the possibility of 1 person throwing a rising 90mph fastball out of 100 people is probably zero. But if each person were capable of throwing a 100mph baseball 15,000 times, I would expect some of those pitches to rise above the normal parabolic anomaly.
Even if I am wrong, it would be nice to have someone else discuss the physical possibility of a rising fastball.
Posted by: Eric | December 6, 2009 11:34 PM
Everyone here seems to be confusing a hypthetical model illustrating the fundamental reality of gravity vs. reality. That a bullet fired in normal air (a spinning bullet for that matter) questionably horizontal, was within 10% of the time hitting the ground as an object dropped vertically, is proof positive that an object with horizontal speed falls at the same speed as an object dropped vertically, with no relative speed. I think that any joe on the street would assume that the difference in time would have be much greater. Expecting a perfect result under very imperfect circumstances is ludicrous.
Posted by: Mark Miles | January 4, 2010 4:53 PM