If a person falling through the air assumed a position to cause the most possible drag, how long....?

If you want to know why the answers to your questions are what they are, I recommend you read the wiki article at Source 1.

Short answer:

"Based on wind resistance, the terminal velocity of a skydiver in a belly-to-earth (i.e.:face down) free-fall position is about 195 km/h (122 mph or 54 m/s). . . . a speed of 50% of terminal velocity is reached after only about 3 seconds, while it takes 8 seconds to reach 90%, 15 seconds to reach 99% and so on." -- Source 2

Assuming the average man's mass is 80kg (source 3), I come up with around 600m (about 2000 feet) to reach 99% of terminal velocity.

Longer Answer:

I used this formula for terminal velocity Vt from Source 1:

(1) Vt = √ (m * g / k),

where:

m = the average mass of a man = 80kg

g = accel of gravity

k = ½ * ρ * A * Cd

to figure out the value of k using the numbers from the Short Answer above. I got:

(2) k = 0.27.

Then I took the equation for velocity v in Source 4:

(3) v = √ (m * g / k) * tanh ( √ ( k * t / m) )

and graphed it using Source 5. Then I summed up the area under the velocity curve from t = [0, 15] and got ~ 600m for the distance to reach 99% of Vt (requiring 15s).

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