Apropos of this comic, the other night I was watching the “Mythbusters” episode where they test being talked through landing a commercial airplane (plausible, it is possible to be talked down but the odds against circumstances that would lead to this are astronomical) and three parachuting “myths” all from one scene in “Point Break”. To test the length of the scene (about 90 seconds before the chute is opened) they tossed Buster out of a plane at 4000 feet (the altitude of the plane when Patrick Swayze jumped). Buster hit the ground in 31 seconds. For some reason this caused me to go get paper and pencil and start doing so calculations.
At first I couldn’t figure out why it took so long, then I remember that Buster was not falling in a vacuum. Then I did the calculations based on a terminal velocity of 120 feet per second (the number they calculated as the terminal velocity of a person in the “spread eagle” position) and Buster fell too fast. (Because being a crash test dummy he didn’t know that he could live a couple seconds longer if he’d kept his arms out wide.) So I decided to find out what Buster’s actual terminal velocity was.
Here’s where I need someone to check my math. (Sign convention: plane is at x=0 and positive x is towards the ground. This gets rid of a bunch of superfluous negative signs)
We know that:
xt= x0 + v0(t) + ½ a(t2) vt = v0 + a(t)
If:
xa = distance traveled while accelerating,
xt = distance traveled while at terminal velocity,
ta = time accelerating, and
tt = time at terminal velocity
vt = terminal velocity
a = 32ft/s2, the acceleration due to gravity
Then:
xa + xt = 4000ft,
ta + tt = 31s
And:
xa = ½(32ft/s2)( ta2),
xt = vt (tt), and
vt = (32 ft/s2)( ta)
So:
tt = 31s - ta
xt = (32 ft/s2ta)(31s- ta)
xa = ½(32ft/s2)( ta2)
(let’s drop some units to make this easier to read)
4000 = [(32 ta)(31- ta)] + ½(32)( ta2)
4000 = [(32 ta)(31) – 32ta]+ (16ta2)
4000/16 = [(32)(31ta) – 32ta2]/16+ [(16)( ta2)/16]
250 = [(62ta) – 2ta2] + [ta2]
250 = 62ta –ta2
0 = 62ta – ta2 – 250
(Rearrange into the standard order)
-ta2 + 62ta – 250 = 0
Use the quadratic equation to find ta
if ax2+bx+c=0, then x=[-b±(b2-4ac)½] /2a
ta = [-(62) ± ((62)2 – 4(-1)(-250))1/2]/2(-1)ta = [ -62± (3844 -1000)1/2]/-2
ta = [-62 ± (2844)1/2]/-2
ta = [-62 ± (53.3292)]/-2
ta = [-62 + (53.3292)]/-2, [-62 – (53.3292)]/-2
ta = (-8.6708)/-2, (-115.3291)/-2
ta = 4.3354s, 57.66
(57 seconds is longer than the whole fall took, so it can’t be the time spent accelerating)
ta = 4.3354s
vt = (32 ft/s2)( ta) = (32 ft/s2)(4.3354s) = 138.7328 ft/s
Buster’s terminal velocity was 139 ft/s
We can check this by finding the distances he fell while accelerating and while at terminal velocity and making sure they add up to 4000ft.
tt = 31s – ta = 31s – 4.3354s = 26.6646s
xa = ½(32ft/s2)(ta2) =½(32ft/s2)( 4.3354s2) = 300.7311ft
xt = vt (tt) = (138.7328ft/s)(26.6646s) = 3699.2546ft
xa + xt = 300.7311ft + 3699.2546ft = 3999.9857 feet
Not quite but close enough for government work.;)
(Actually, four decimal places is far more precise than my initial inputs, so my answer is way more precise than my inputs. Actually, I’m a huge geek. And typing math is a pain.)
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