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:

x

_{t}= x

_{0}+ v

_{0}(t) + ½ a(t

^{2}) v

_{t}= v

_{0}+ a(t)

If:

x

_{a}= distance traveled while accelerating,

x

_{t}= distance traveled while at terminal velocity,

t

_{a}= time accelerating, and

t

_{t}= time at terminal velocity

v

_{t}= terminal velocity

a = 32ft/s

^{2}, the acceleration due to gravity

Then:

x

_{a}+ x

_{t}= 4000ft,

t

_{a}+ t

_{t}= 31s

And:

x

_{a}= ½(32ft/s

^{2})( t

_{a}

^{2}),

x

_{t}= v

_{t}(t

_{t}), and

v

_{t}= (32 ft/s

^{2})( t

_{a})

So:

t

_{t}= 31s - t

_{a}

x

_{t}= (32 ft/s

^{2}t

_{a})(31s- t

_{a})

x

_{a}= ½(32ft/s

^{2})( t

_{a}

^{2})

(let’s drop some units to make this easier to read)

4000 = [(32 t

_{a})(31- t

_{a})] + ½(32)( t

_{a}

^{2})

4000 = [(32 t

_{a})(31) – 32t

_{a}]+ (16t

_{a}

^{2})

4000/16 = [(32)(31t

_{a}) – 32t

_{a}

^{2}]/16+ [(16)( t

_{a}

^{2})/16]

250 = [(62t

_{a}) – 2t

_{a}

^{2}] + [t

_{a}

^{2}]

250 = 62t

_{a}–t

_{a}

^{2}

0 = 62t

_{a}– t

_{a}

^{2}– 250

(Rearrange into the standard order)

-t

_{a}

^{2}+ 62t

_{a}– 250 = 0

Use the quadratic equation to find t

_{a}if ax

^{2}+bx+c=0, then x=[-b±(b

^{2}-4ac)

^{½}] /2a

t

_{a}= [-(62) ± ((62)

^{2}– 4(-1)(-250))

^{1/2}]/2(-1)t

_{a}= [ -62± (3844 -1000)

^{1/2}]/-2

t

_{a}= [-62 ± (2844)

^{1/2}]/-2

t

_{a}= [-62 ± (53.3292)]/-2

t

_{a}= [-62 + (53.3292)]/-2, [-62 – (53.3292)]/-2

t

_{a}= (-8.6708)/-2, (-115.3291)/-2

t

_{a}= 4.3354s, 57.66

(57 seconds is longer than the whole fall took, so it can’t be the time spent accelerating)

t

_{a}= 4.3354s

v

_{t}= (32 ft/s

^{2})( t

_{a}) = (32 ft/s

^{2})(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.

t

_{t}= 31s – t

_{a}= 31s – 4.3354s = 26.6646s

x

_{a}= ½(32ft/s

^{2})(t

_{a}

^{2}) =½(32ft/s

^{2})( 4.3354s

^{2}) = 300.7311ft

x

_{t}= v

_{t}(t

_{t}) = (138.7328ft/s)(26.6646s) = 3699.2546ft

x

_{a}+ x

_{t}= 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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