On the popular Mythbusters TV show, one myth to be tested was that in World War II, a gunner fell out of an airplane without a parachute from 22,000 feet, and that he survived because his fall was partially softened by the shock wave from an exploding bomb on the ground just prior to his own impact.

To test the plausibility they dropped a dummy from a height they thought was sufficient to result in the dummy's reaching "terminal velocity", the velocity at which air resistance exactly balances the acceleration of gravity. It's assumed that terminal velocity for a falling human is 120 miles per hour. Co-host Adam did some calculations and determined that terminal velocity would be acquired in 5.5 seconds and require only 500 feet of falling.

What's wrong with the calculation? Assume that the acceleration (call it deceleration if you like) due to air resistance at any given instant is proportional to the square of the downward velocity.

How fast would the falling body be traveling after freely falling 500 feet subject only to gravity and air resistance? How far would a body have to fall to get within 1 mile per hour of terminal velocity? (Again assume terminal velocity to be 120 miles per hour.)

(In reply to Numerical solution - Part II by Kenny M)

here is a closed form solution.  We want to solve
v'=g-A v^2
  with g=32.17 ft/s^2
given that the terminal velocy is 120 mph, (176 ft/s),
  A=32.17/176^2  ft^-1

The trick is to work directly with the velociy.  We'll integrate
1/(g-A v^2)  dv = dt
with the initial condition v=0 at t=0.

t=1/g  ∫1/(1-(A/g)v^2) dv   {Integral goes from 0 to vfinal}

Quick change of variables, x = Sqrt[A/g] v
  but Sqrt[A/g]=1/(176 ft/s)=1/vt where vt is the terminal velocity.
  so x=v/vt, and dv = vt dx

t=vt/g ∫ 1/(1-x^2) dx  {Integrate from 0 to vfinal/vt}

The integral of 1/(1-x^2) is 1/2 (ln(1+x)-ln(1-x))  (As long as x<1)

We want v to be 119 mph, which is 119/120 176 ft/s.  So the answer is
t=vt/(2 g)(ln(1+119/120)-ln(1-119/120))
 = 14.98

Posted by eebo on 2007-02-20 17:24:46