# Physics Question: Jumpin outta planes

If i jumped out of a plane at 13,500 ft (2.5 miles) and freefalled for 60s how fast would i be travelling at the end of the freefall in miles/hour? This is all ignoring wind resistance. I know the accerlation of gravity is 32ft/s/s but how do you work that into m/hr? Annndd finalllly, what, if any, would by the the g-force at the end of the freefall?

Freefalling with no air resistance for 60 seconds, you’ll be doing over 1300mph and have covered over 50,000 feet.

If you started just 13,500’ over the ground, then you actually fell for about 29 seconds and reached a speed of about 635 mph. The deceleration when you hit the ground depends on the distance in which you stopped, which depends on what you hit. We can be fairly confident it was severe.

It might be worth wondering how, if there is no air to provide resistance, that plane of yours managed to reach 13,500’.

how bout the G-Force?

Velocity^2=Vinitial^2+2(Gravity)(DeltaX)
Fg=[(MassEarth*MassObject)/R^2]G

F = m*a; Force == mass times acceleration

Work == F*d == mgh
(This is a simplified conservation of energy expression, where I assume that all of the gravitational potential energy of being up in the plane is transformed into speed, then is absorbed in stopping you at the ground.

This simplifies to:

mgh/d == F

Where g is a constant relating to the earth’s gravitational pull on you. Looking up, I see you know it’s 32 ft/s^2, but I’d rather call it 9.8 m/s^2. Call me old-fashioned.

We need your weight (I’ll assume you weigh about 165 lbs or 75 kilograms), the height of the plane (13500 feet == 4115 metres) and, like Xema said, the distance you penetrate into the ground. I’m going to assume you hit concrete (which, for the keeners, has a modulus of elasticity in the many megapascals, so the ‘d’ is almost entirely coming from the motion of his Centre of Mass relative to the ground as he flattens in the impact). I’m guessing you’re about 1.8 metres == 5 foot eleven, and that you’ll hit feet-first, to reduce the force (though this means you hit with a small surface area. I expect your legs will shatter).

Anyway, this gives a d of about 0.9 metres.

mgh/d == F
(0.75)(9.8)(4115)/0.9 == F
33605.8 == F

So, the ground exerts 34 kiloNewtons on you.

Normally, the earth exerts 9.8 Newtons on each kilogram of you… if you weigh 75 kilograms (165 lbs), that’s 735 Newtons.

34 kN/735 N == 45.7

So, The force you feel is about “48g”. Keep in mind this is a very rough estimate, where I’ve assumed:

Deformation of the ground is negligible
You hit feet-first. Or head-first. But not sideways. (affects d)
You weigh 165 lbs
No energy is lost on the way down to friction or other things.

If I understand it correctly, you would be experiencing exactly one G, because the only force acting on you would be in fact, the gravity of the earth. With air, you’d have a net force of less than one G, because the air resistance would be pushing in the opposite direction.

When you hit the ground, even at only a couple hundred miles an hour, you’d experience a LOT more Gs. I think the Guineess book record for survived G force is a guy who survived a bad parachute jump.

OK, I’m going to dash off some calculations before going to the grocery store; if I’ve made any idiotic errors or if any of my assumptions are stupid or otherwise invalid, no doubt I will be corrected by others.

So, let’s assume the following: 1) that you land on your feet, 2) that the ground is unyielding, or that the yield is negligible, so the only thing decelerating you is your body’s impact, 3) that your leg bones shatter in a linear manner (which they wouldn’t, but hey, we’re simplifying here), 4) that they collapse up to, say, mid-thigh, a total distance of three feet (for further simplification), and 5) Xema’s calculations are correct and, after you jump out of a plane at 13500 ft with no air to help you reach a terminal velocity, you hit the ground traveling 635mph. (I haven’t verified that, and I’m in a bit of a hurry so I’m not going to.)

So with all those assumptions, your vital organs, especially your brain, are going to accelerate (negatively) from 635mph to 0mph over a distance of three feet. Acceleration (a) is related to velocity (v), time (t), and distance (d) by the equations v=at and d=vt, so:

t = d / v

t = (3 ft) / (635mph)

Converting miles to feet and hours to seconds, we get:

t = (3 ft) / (3352800 ft / 3600 s)

t = (3 ft) / (931.3 ft/s)

t = 0.00322 s

So assuming a linear deceleration, which was assumption 3 above, it takes 0.00322 seconds to go from 635mph to 0mph over a distance of 3 feet. Now, since d=vt, and v=at, we can do this:

d = vt

d = (at)t

d = at[sup]2[/sup]

a = d / t[sup]2[/sup]

a = (3 ft) / ( (0.00322 s)[sup]2[/sup])

a = (3 ft) / ( 0.00001037 s[sup]2[/sup])

a = 289296 ft/s[sup]2[/sup]

Converting feet to meters, we get …

a = 88177 m/s[sup]2[/sup]

Since acceleration from gravity is normally 9.8m/s[sup]2[/sup], we can calculate the ratio of our massive acceleration to gravity’s acceleration to get the gee forces:

g = a / a[sub]g[/sub]

g = (88177 m/s[sup]2[/sup]) / (9.8 m/s[sup]2[/sup])

g = 8997.7

So, if all our assumptions hold, and if I haven’t made any boneheaded errors, you would experience very close to 9000 gees.

On preview, I see that wolfstu came up with a much smaller answer of about 48g. What the crunk? I guess I did make a bonehead error (or twenty) somewhere in there … but for now, it’s off to the grocery store before my wife brutally slays me, so I’ll let the SDMB physics deities correct me whilst I comparison various brands of mayo.

Ah, no, chorpler I screwed up. I assumed his mass was 75 kg, but for some reason I used 0.75 in my calculations. I must be tired…

That just scales my answer by 100…

3360583.3 Newtons == 3.4 MN

We can do a few more fun things, while we’re at it:

Pressure on your feet when you hit:

Pressure == Force / Area

Where the area is the amount of floor space occupied by your feet when you stand. Roughly. Let’s say your feet are about my size (North American Mens 12). Roughly with my ruler, I’d say they cover, oh, 0.0375 m^2. That’s about 58 square inches.

That works out to a pressure of just about 89 MegaPascals. Like, nine hundred times the ordianary pressure of the atmosphere on your skin.

Energy Released
We can also estimate that the work done in stopping you (the amount of energy expended) is just W == mgh == 3024525 Joules, which could run a hundred-Watt bulb for about eight and a half hours, or boil about 8.9 kilograms of room-termperature water (two and a half gallons or so). Of course, instead, the energy went into destroying your body and the sidewalk.
Let’s look up the modulus of elasticity for human bone, shall we? (Google pause). This site says it’s about 9.8 GPa. Using Hooke’s law (integrating it for the strain-energy) and the diameter of a femur, I’m guessing your thighs would get about 0.0248 metres shorter… before they snapped. Oh, and would they ever snap. The stress on them would be 2.9 GPa, about 17 times what they could take if you hit just right. Which you wouldn’t, since all the rest of the body parts supporting them are pretty bendy, and would more than likely fail first, absorbing some of the energy.

You can be pretty confident you wouldn’t walk away fro this landing.

OK, a friend of mine looking over my calculations above points out that in the equation d=vt, v needs to be the average velocity, which is (at)/2 instead of just at. So that halves my answer of 8997 gees, making it 4499 gees, much closer to wolfstu’s answer.