Do you ever have one of those days where you start wondering about some esoteric little problem and can’t let go of it until you find the answer? It can be anything. How many angels can dance on the head of a pin? How many chips are in a bag of Fritos? You know…stupid stuff that won’t matter in the immediate future much less a year or a decade from now. I had one of those days last Sunday.
I had to make a quick trip down to Centerville, Texas to meet a guy who was coming up from Houston for a little business we had to conduct together face to face. It’s a 2 hour drive from Castle Erickson to Woody’s Smokehouse in Centerville. I was a little tired from staying up too late on Saturday night, and I needed a distraction to keep my mind active while on the drive.
So, what little esoteric nugget do I come up with? Something profound? No. Something amusing? Well, that depends on whether or not simple math and basic physics are your idea of a knee slapper.
It all started with a MythBusters episode I saw a little while back where Adam and Jamie were trying to determine whether a bullet fired straight up (90 degrees perpendicular to the ground) could cause a fatal injury when falling back to earth. Now the MythBusters conclusively proved that a bullet fired straight up at 90 degrees would fly straight up and fall back down at a terminal velocity of something like 100 miles per hour. Since the atmosphere is not a perfect, frictionless environment, the projectile would eventually stop spinning allowing the bullet to destabilize and start stumbling.
Add to that little memory nugget the fact that I’ve been trying to educate myself about all aspects of the shooting sports…including ballistics. So, I decided to see if I could figure out how high a 230 grain bullet fired straight up at 1000 feet per second would theoretically travel before coming back home to papa.
Now, before I go on, allow me to say that I haven’t had a math or physics class in about 18 or 19 years after getting a C in Calculus I on the second try and an A in Analytical Geometry on the third try (yeah, I’m stubborn that way sometimes). That right there pretty much killed any desire or hopes I had for pursuing a degree in engineering.
Anyway, the math necessary to figure out the answer to my problem is deceptively simple once you understand the physics part of the equation. Some people might think that there is not enough information provided above to figure out the answer, and they would be partially correct. You do have to know what the force of gravity is to get the answer. Other than that, it boils down to straight forward multiplication, division, addition and subtraction.
I’ll help you out here and clue you in that the force of gravity is roughly 32 feet per second squared (32.163 ft/sec sq. to be exact…remember that figure…it’s necessary if you want to figure out muzzle energy on your own without a ballistics table).
So, we point our theoretical pistol straight up so the barrel is 90 degrees relative to the ground and pull the trigger. Our theoretical 230 grain bullet leaves the muzzle at an initial velocity of 1000 feet per second. As a quick aside here, there are 7000 grains in a pound making a 230 grain bullet 0.5257 ounces or 0.03285714 pounds. Not considering friction (since I didn’t have a reliable source handy from which to obtain figures), the only force acting on the bullet once it leaves the barrel is gravity. Gravity is Decelerating the bullet at a constant rate of 32 feet per second squared. In other words, every second the bullet is in flight, it slows down by 32 feet per second.
Now, with this bit of information, we can arrive at the first interim step along the way to answering our original question: how much time will the bullet be in flight before running out of steam. Basic long division is our friend here. 1000 feet per second divided by 32 feet per second squared leaves us with 31.25 seconds of travel time.
That answer allows us to take the next step in answering the problem. We now know that we need to knock 32 feet per second off the initial velocity and each subsequent velocity 31.25 times. Each of those 31.25 speeds represents the distance that the bullet would have traveled in each of those 31.25 seconds. This is where we break out elementary addition and subtraction. 1000 feet per second minus 32 feet per second equals 968 feet per second. 968 minus 32 equals 936. And so on and so forth. Eventually, 31.25 times later we get to the last quarter second of the bullet’s flight where it expends the last 8 feet per second of its velocity (notice the symmetry that 8 is ¼ of 32) to ever so momentarily hover in midair at a relative speed of exactly zero feet per second just before starting its return trip to terra firma. If you add up each of those recorded velocities, you come up with a total distance traveled in 31.25 seconds of 16,128 feet. Dividing that figure by 5280 feet per mile, we learn that the maximum potential altitude for a 230 grain bullet (any weight bullet for that matter) fired at 1000 feet per second is 3.05454… miles.
I know that some more seriously minded math and science types might quibble that there are “simpler” and more precise ways to calculate the answer to that question. Possibly. A quick Google search from the comfort of home using my DSL connected laptop instead of my 3G iPhone while traveling 70 miles per hour on I-45 suggest that I use the formula “distance = initial velocity plus ½ the rate of acceleration times the square of the amount of time involved.” That would look like “d = 1000 ft/sec + (1/2 * 32 ft/sec * 31.25 sec * 31.25 sec)” or 16,625 feet (3.148674 miles).
So, all things considered, I don’t think I did too badly for a tired guy driving a car and fiddling with a phone and a pad of paper. A difference of 500 feet (or just a touch less than 1/10 of a mile) is a pretty good swag if I do say so myself.
Now, for the bonus question: if the theoretical 230 grain bullet reached its maximum altitude and is struck by a Boeing 737 traveling at an airspeed of 500 miles per hour at the very instant that the bullet’s velocity reached zero, how much kinetic energy imparted to the airplane by the impact with the bullet?
I’ll let you all noodle that one for a bit. If you’re nice, I might even post the answer.
