A physics question, courtesy of Arnold Schwarzenegger

October 28, 2022 1:17 AM

There's a scene in the movie Commando in which our hero, played by AS, jumps out of a plane shortly after takeoff and falls, sans parachute, probably 100 feet or so into some marshy shallows. When he hits the water, he goes straight in, ie, no forward momentum sending him skidding along the surface. My question are: is this an accurate depiction of the physics at work? Would forward momentum have dissipated in the time it took Arnold to fall from a great height?
posted by BadgerDoctor to Science & Nature (13 answers total)

It'll depend on the takeoff speed. If you mention what kind of plane it was, we could probably make a rough estimate.

Also note that planes tend to take off into the wind, so when Arnie falls out of the plane he'll catch the wind which will tend to slow him down.
posted by Johnny Assay at 4:24 AM on October 28, 2022 [1 favorite]

In a case like this any forward momentum can only be dissipated by air resistance. It takes about 2.5 seconds to fall 100ft/30m starting with zero vertical speed. Adding some upward speed because of the plane just having taken off it would take maybe four seconds.

Air resistance depends on a few factors: posture and loose or tight clothing, but you're not going to scrub a plane's takeoff speed down to (near) zero in just four seconds.
posted by Stoneshop at 4:28 AM on October 28, 2022 [1 favorite]

Possible? Yes, via slow forward speed and large upward speed of plane, strong headwind, Arnie jumping tailward and making himself large in horizontal cross section.

Likely? No. Except likely the stunt people dgaf about making the forward momentum look right.
posted by SaltySalticid at 4:57 AM on October 28, 2022

Best answer: The scene in question
posted by Gorgik at 5:03 AM on October 28, 2022 [3 favorites]

Yep, all spherical cow not like.... forward momentum will decrease rapidly because there is no force to sustain forward momentum versus air resistance. On the down side, gravity will increase your speed. Going all vector on it the horizontal will get smaller and smaller and the vertical will get larger and larger (up to terminal velocity) and end up being closer to vertical.

If you toss something out the window of your car does it go whoosh and heads backwards? It's not a zero drag coefficient object and has no forces acting horizontally only gravity acting downwards. It will slow and drop. Schwarzenegger is a big dude, lots of cross sectional area, high drag coefficient.

Ehhhh... not really enough information. :)
posted by zengargoyle at 5:12 AM on October 28, 2022

Best answer: So the question is, essentially, does the x-velocity of a falling object reach zero before the y-position reaches zero. Here's how to solve that:

If you draw a vector diagram on Arnie, you'll see two forces acting on him as soon as he jumps out of the plane: gravity (-y) and air resistance (??). Doesn't matter which direction air resistance actually works because we can break it into its orthogonal components, so let's say you have gravity (-y), vertical drag (+y), and horizontal drag (-x).

Parasitic drag (which is what we're working with here) is defined by the equation D = 0.5 * rho * U^2 * A * Cd, where-
-rho is the air density
-U is the airspeed
-A is the projected area, or the area of the body that the airstream "sees"
-Cd is a drag coefficient determined by the geometry of the body.

You'll notice that the drag forces acting on Arnie are proportional to the square of the velocity. Now let's look at how we get a velocity out of this. Remember Physics 1: F = m * a. We want velocity, but our equation tells us acceleration! We need to do some calculus:

a = dU/dt (remember we've defined U as airspeed here)
DU/dt = F / m
U = integral (F / m) dt

Since F (the sum of all of our forces in one orthogonal direction) is simply our horizontal D-value, then

U = integral [ (1/2m) * rho * U^2 * A * Cd] dt

This integral has limits from 0 (time immediately jumping out of the plane) where U0 = forward speed of the airplane to some unknown time t1 where U1 = 0. Everything inside of that integral is a constant except for U, and this is where we're going to run into a problem. Because U is also a function of t. Really, you have this:

U(t) = (rho * A * Cd) / (2m) * integral [U(t)^2] dt

...and if you remember more calculus than I do, you might be able to solve that equation for t1. (This is where my TAs in college would declare that the solution is "intuitively obvious.") So, at this point you have the time it would take for Arnie to slow down to a zero forward velocity. Is that number bigger or smaller than the time it takes him to drop to the ground? At this point, I would assume that he hits terminal (vertical) velocity instantaneously, and you can use the following equation to figure that value out:

Vt = sqrt [ (2 * m * g) / (rho * A * Cd)]

But note! A and Cd here are different than A and Cd in the forward velocity question. These would relate to the projected area of Arnie looking up at the soles of his shoes (I'm also assuming here he jumped out of the plane like a pencil and is not moving). Anyway, you can make some assumptions about what A and Cd are for each equation based on some easily available reference material (make it easy on yourself and assume he's a large rectangular block, so Cd is 1 in both axes).

Then it's just simple math to determine how long he takes to hit the ground based on how far he has to travel, i.e. how high the plane was when he jumped out. Compare the two times, and if t1 from the horizontal equation is less than the time spent in free fall then you can rest assured that he's lost all forward momentum before hitting the water.
posted by backseatpilot at 5:22 AM on October 28, 2022 [5 favorites]

Also, theoretically, the air around the plane is compressed and denser and moving faster because the plane is squeezing it out of the way while it's pushing itself through the sky. So that initial let go backwards push is going to be rather strong.
posted by zengargoyle at 5:23 AM on October 28, 2022

ssume he's a large rectangular block, so Cd is 1 in both axes)

He's a monolith, much wider from the front than the bottom, wider than he is thick. Arms can be assumed round and contribute the same depending on position. Legs will be full on from the front and reduced from the bottom unless he's doing a split. Chest is at least twice as wide as it is thick. Head only counts forwards because downwards it's behind the bottom stuff. The cross section is vastly different.
posted by zengargoyle at 5:40 AM on October 28, 2022

Well, evidently Arnie is already going slower than the plane he's holding on to as his jacket is just flapping as in a moderate breeze, not the takeoff speed of a Tristar, 180 knots (210 miles/hr, 330 km/hr)
posted by Stoneshop at 5:43 AM on October 28, 2022 [4 favorites]

That is say from the front he's 6x2 and from the bottom he's 2x1.
posted by zengargoyle at 5:44 AM on October 28, 2022 [1 favorite]

That's the A in the equation. The Cd is how efficiently he is penetrating the air (eg think about the decreased air resistance of adding a pointy front to a flat box, same area but the latter has reduced Cd). Arnie might actually have a higher more resistive Cd than a square box in both directions because of the flappy jacket he is wearing.
posted by Mitheral at 6:11 AM on October 28, 2022

Best answer: As Arnie moves through the air he will be slowed down by air resistance (drag). This will decrease the horizontal component of his speed.

There are several factors that relate to the amount of air resistance:

a) Surface to volume ratio. The larger the surface to volume ratio, the more air resistance will affect the object. The shape affects the drag coefficient, a badminton shuttle will have more air resistance than a golf ball.
b) The surface of the object. If the surface is rough, the air resistance will be greater.
c) Speed. As speed increases, so does air resistance. Drag force is proportional to the square of the speed of the object.
d) Mass. The smaller the mass of an object, the more air resistance will affect it. For example a feather, compared to a stone.

If we assume the aircraft is travelling at 155 mph or 250 km/h, Arnie has a drag coefficient of 1 and a surface area of 1.9 m^2 (all finger in the air guesstimates)
we can plug those numbers into the formula:
Fd = cd 1/2 ρ v2 A
Giving a force of 5497 N at the point where he exits the aircraft.
As he slows down the force will reduce, so at a speed of 77 mph the force is only 1374 N.

The next part to calculate is how long it will take to decelerate horizontally, using the impulse momentum theorem:
F(∆t) = m(∆v)
155 mph or 250 km/h = 69.44 m/s (rough estimate of takeoff speed)
5497 = 100 kg (69.44 - 0 m/s) / ∆t
solving for (∆t) = 1.26 seconds

But that is assuming we have a constant force which is not going to be the case as he slows down.
Taking the force at an average horizontal speed of 77 mph (which is probably wrong but gives us a rough ballpark)
F(∆t) = m(∆v)
1374.42 (∆t) = 100 kg (69.44 - 0 m/s)
solving for (∆t) = 5.05 seconds

So now all thats left is to calculate how long it takes to fall 100ft, if that is less than 5.05 seconds then he will still be travelling horizontally on impact.

Using this free-fall calculator
falling from 100ft = 2.49 secs
falling from 400ft = 4.9 secs

There is surely a more accurate way to calculate this, but my very rough estimate is that he would need to exit the aircraft at around 500 feet or higher to then hit the water vertically.

The Lockheed TriStar has a climb speed of 14.3 m/s (2820 ft/min) so it would take 5.6 seconds to reach 500 ft.
In the movie Arnie jumps out 11 seconds after take off, so if thats accurate he would have been nearer 1000 feet.

So the answer is that Yes Arnie could have landed with little horizontal speed, but he would have done this from around 10 times the highest possible survivable height for a fall into water.
posted by Lanark at 6:39 AM on October 28, 2022 [4 favorites]

Ok I watched the clip.
is this an accurate depiction of the physics at work?
I have to take back my initial answer based on description. It is far more non-physical than I had imagined, mostly due to a far bigger/faster plane. This is just typical cartoon physics that do whatever your script wants. Yes due to the good math presented above but it also just looks about as physically accurate as the Coyote hanging in air for a few seconds as he holds up a sign and mugs for the camera before he drops.
posted by SaltySalticid at 10:19 AM on October 28, 2022 [1 favorite]