Question:
After flying what distance will human reach it's terminal velocity (flying flat)?
TX
2010-01-14 08:18:56 UTC
After understanding what "terminal velocity" is and how it works in my previous question, now I have another question. I read at multiple places that terminal velocity is about 200km/h for a human falling with body aligned flat (to minimize speed). I would like to know, after falling what distance, human will hit it's terminal velocity (200km/h). I've googled and found many contradicting answers, so I want to know what is the true answer. So please do not google for it, instead calculate yourself, and if you could give me a formula which only needs object's weight, terminal velocity (human falling flat and falling vertically has different terminal velocities and I'd like to calculate for both of them) and fall point height values as an input - would be great! I know that the formula depends on many other parameters than weight, terminal velocity and height, but please assume "default"("normal") values for them.

SHORTLY: I know that falling from X meters is not worse than falling from Y meters (X being more). I'd like to know what the minimum value of Y is for this statement to be true! Please consider two scenarios if possible : falling flat (hands and legs outstretched), and free falling (accidental fall, unconscious fall, etc.)

Thank you in advance! I need this answer very much!
Four answers:
Randy P
2010-01-14 09:08:24 UTC
If you assume constant-accelerated motion, v^2 = v0^2 + 2ad



200 km/h = 55.6 m/sec



Using the second formula with v = 55.6 m/sec, a = 9.8 m/sec^2 and initial velocity v0 = 0 m/sec gives



d = v^2/(2a) = 158 meters. This is the distance you would free-fall to reach a speed of 55.6 m/sec, assuming constant acceleration of 1 g.



But of course if you are reaching a terminal velocity, then it isn't simply constant acceleration any more. The acceleration is gradually decreasing. You have to solve a differential equation to get the more accurate model of velocity vs. distance or time, and I'm not sure there's a simple formula for the solution. If I were doing this, I'd use a computer.



You could model it on Excel, maybe that will help.



Edited: OK, here's how to do it in Excel.

1. Make four columns for time, distance, velocity and acceleration. I'll assume they're columns A, B, C and D.



2. For initial values put 0, 0, 0, and 9.8



3. In the time column, make even steps in some small increment (0.1 second ought to do) up to say 10 seconds.



4. In the acceleration column, put 9.8 - 0.00317*(previous velocity)^2. That second term accounts for air resistance. That coefficient makes the air resistance = gravity when v = 55.6 m/sec.



5. In the velocity column put v = previous velocity + acceleration * delta-t where delta-t is your time step.



6. In the distance column, put distance = previous distance + velocity * delta-t.



I just wrote a spreadsheet like this. I found that under this model after 10 seconds you've fallen 350 meters, you're moving at 52.5 m/sec and you're still accelerating at about 0.1 g.



At 30 seconds you've gone about 460 meters, you're moving at 54.1 m/sec and your acceleration is down to about 0.05 g.



"Terminal velocity" is an asymptotic value. You don't ever actually get there. So I'm not sure what you'd accept as the answer.
?
2016-10-04 05:02:01 UTC
Terminal Velocity Calculator
Michael
2015-06-24 17:46:27 UTC
The answer is not answerable as it depends on how far away from your natural density layer you are.

The equation Vt=(sqrt( (density object/density medium)-1)*2 is a useful equation to dispaly this.

If you get a table of air density verus hieght from the web you can see that the air density at 30,000 ft is 1.80119E-2 Kg/m cubed and at sea level it is 1.225 Kg/m cubed. Density of a person = 950 Kg/m cubed

Vt = 30,000ft = 490 mph this is not stable and is still accelerationing. At sea level Vt= 119.142 mph

This can be calculated on Excel after copying the table. So the best answer equation needs changing so that the 0.00317 constant is a variable prortional to the density gradient.

What it shows is that when we are in water at a depth equal to our density Vt=0 acceleration =0 Force =0

This is our natural density level caused by our origins.

So from a great height it will take much longer to reach the terminal velocity than from a 1000 ft for example.
Lorenz
2015-03-21 05:29:06 UTC
At 30 seconds you ve gone about 1460 meters!


This content was originally posted on Y! Answers, a Q&A website that shut down in 2021.
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