Terminal Velocity

A falling object, experiencing a general drag force,

FD » bv + cv2,

will accelerate until the drag force, which increases with velocity, becomes equal to the object's weight.  At that instant the net force on the object reaches zero, it ceases to accelerate, and its velocity becomes fixed.  Hence the velocity at which this condition occurs is known as the body's "terminal velocity."

(1) If we assume that the terminal velocity is great enough that the v2 term in the above equation
dominates, then FD » cv2 » ½ρADv2.  Since the condition for terminal velocity to occur is
FD = mg,  we set ½ρADvT2 = mg.  Solving for vT, yields

vT = (2mg/DρA)½.

If r represents the size of the object, then A µ r2, and m µ r3, so vT µ r½, i.e., for an object of
a given density and shape, the bigger the object, the greater the terminal velocity.  Thus a small
insect dropped from the top of the Kimball tower to the ground beneath would probably survive,
a dropped mouse would have a lesser, but still decent survival chance.  A human would have a
very small survival chance and a horse would have virtually no chance of surviving.

(The dependence of terminal velocity on size is well illustrated by atmospheric water droplets as
in the following table:

 Terminal Velocities of Different Size Droplets (and Particles) Diameter [μm] Terminal Velocity Type of Droplet (Particle) [m/s] [ft/s] 0.2 0.000 000 1 0.000 000 3 (Condensation nucleus) 20 0.01 0.03 Typical cloud droplet (mist) 100 0.27 0.9 Large cloud droplet (drizzle) 200 0.70 2.3 1000 4.0 13.1 Small raindrop 2000 6.5 21.4 Typical raindrop 5000 9.0* 29.5* Large raindrop

*We can be thankful for reasonable terminal velocities.  Were it not for the braking effects of air
resistance, a raindrop falling from 10 000 feet above the surface (such fall distances are
common in thunderstorms) would hit the ground at about 800 ft/s = 244 m/s = 545 mph.  With
a barrage of large droplets at such velocities, exposure to thunderstorms would be lethal.

(2) At low relative speeds the term bv dominates the drag force.  An object falling in a liquid will
achieve terminal velocity at a speed low enough that the cv2 term in the drag force is negligibly
small.  Thus we can write bv = mg, or

vT = mg/b.

In this case, since ma = m dv/dt = Fnet = mg - bv,

dv/dt = - bv/m + g.

This is a first order differential equation.  If we assume the initial condition v(t = 0) = 0, having
taken downward as the positive direction, the solution is

v = (mg/b)(1 - e-bt/m), or

v = vT (1 - e-t/τ),

where we have inserted the time constant τ, defined τ º m/b, which is the amount of time it takes
for the difference between the velocity and the terminal velocity to shrink to 1/e = 36.788% of its
initial value.

(What is the terminal velocity of a falling human?  According to various sources a "balled-up skydiver falls at about 200 mph, but with arms and legs extended, that is reduced to about 125 mph.  On the other hand, well-practiced stunt skydivers in stand-up or diving postures have achieved speeds up to about 320 mph.)

Precisely speaking, the terminal velocity of an object is slightly greater at high elevations than at low elevations because the resistive fluid (air) is denser at low elevations.  For the same reason, terminal velocity increases slightly with increasing humidity.