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.)

Part I:

Adam's calculations used an air resistance force that is too low relative to the assumptions.

 

Although my differential equations are very rusty, I submit the following argument:

For NO air resistance, the increase in speed of fall is determined solely by gravity, and an energy balance will suffice.

 

Delta Kinetic Energy (KE)  = Delta Potential Energy (PE)

0.5 * m * v^2 = m * g * h

m = mass of Buster (assumed to be 150 lbs/32.3 ft/s/s)

v = falling velocity ft/s

g = gravity constant = 32.2 ft/s/s

h = height fallen, in ft

 

For this case (no air resistance) 120 mph (= 176 ft/s) will be achieved in 487 ft, which so far eliminates nothing.

 

Expanding the equation to include loss from air resistance:

 

KE = PE – (work done by air resistance = Wair)

Using Adam's numbers

 

.5 * (150/32.2) * 176^2 = 150 * 500 - Wair,    Wair ~ 2851 lb-ft.

 

Work = Integral (Force * differential displacement), and in this case force is not constant, but is a squared function of velocity.  Noting that at terminal velocity (there is no acceleration at terminal velocity) the air resistance force = the weight of the dummy = 150 lbs, then the air resistance force, Fair = .004842 * v^2  lbs.

 

The average air resistance force over the period of falling from 0 ft/s to 176 ft/s is one third the maximum force (average value = area under the curve divided by the interval, = 1/3 the max for a parabola) = 50lbs.

 

However, going back to the work done according to Adam, Wair, the average force, during that time is 2851/500 = 5.7 lbs.  Therefore Adam only accounted for about 10% of the air resistance force in his calculations.

 

For Part II:

 

The equation of motion (summing forces) is:

 

Weight – air resistance force = mass * acceleration    or

150 - .004842 * v^2 = (150/32.2) * dv/dt , a differential equation in v & t.  As I said, my math here is rusty, so I'll let some one else solve it.

Edited on December 27, 2006, 10:44 am

Posted by Kenny M on 2006-12-27 10:42:26