CBR Revenge Fic

I wrote this snippet last night to blow off some steam after dealing with a particularly annoying fanboy on the Rumbles board yesterday. The idiot was making some rather ridiculous comments on a LOTR characters vs. Conan thread, including statements such as Conan can resist ANY magics thrown his way via sheer willpower (after which Munchy asked 'how do you resist a magic hand grenade by sheer will?') and the claim which started this particular revenge-fic: "No swordsman can stand versus the Cimmerian."

Needless to say, the urge to start a match with Conan vs. Lina Inverse or Kenshin was quite strong. But to save the board the agravation and the display of pettiness, I think I'll keep my little piece of vengeance on my journal.

( read the fic )

Because it won't leave me alone...

...and I have a sneaking suspicion that I might need it in the future when Jared tries to disprove the laws of physics or mathematics, again.

The equation for terminal velocity
Drag force: D = Cd * rho * v ^2 * A / 2
v = sqrt ( (2 * W) / (Cd * rho * A) )

Where Cd is the drag coefficient. (Let's assume 0.5 for simplicity.)

rho is the air density; let's use 1.2854 at -500 m.

Now, we know the terminal velocity of a skydiver without a parachute is about 55 m/s. Let's assume that's the end speed here.

Let's also assume that said skydiver is about 180 cm tall (almost 6') and 50 cm across (about 20") for a frontal area of 9000 cm^2 = 0.9 m^2. Let's also assume the guy has a mass of 80 kg (about 176 lbs).

So, since terminal velocity is the point at which weight = drag, my dilemma is just exactly when a person in free fall hits that point. More importantly, how that would be calculated...

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The problem, of course, is that the acceleration isn't constant, which means the major equations of kinematics are useless.

I suppose one way would be to assume an *average* acceleration of 9.8/2 kg*m/s^2... in which case, starting from rest, we have v = at, or 55 = 9.8t/2, where t is about 11 s...

(tbc)