# Q & A: Reaching terminal velocity.

Q:
what distance fallen is required for a person to reach terminal velocity
- Paul
Lewisville, NC
A:
You can't answer that question exactly because the terminal velocity depends on many variables that change from situation to situation.  Terminal velocity comes about when the frictional forces of the air are equal to the force of gravity.  The frictional force of air depends on air pressure and density, shape of the falling  body, etc.   For example the terminal velocity of a sheet of paper or a balloon is much smaller than the terminal of a BB weighing the same amount.

However, you can make an estimate for a human body.  It is said, although I've never tried it myself, that the terminal velocity of an ordinary person falling in normal atmosphere at low altitudes is around V0 = 200 kilometers per hour ( ~ 120 mph).  Using the standard equations of motion and assuming that the air resistance force is proportional to the velocity squared then you can solve for the velocity and distance.  There are two parameters in the solution in addition to V0:  the characteristic time,
T0 = V0/g =  5.6 sec, and the characteristic distance,  X0 = V0T0 = 315 m.
The full solution is V = V0 tanh(T/T0)   and X = X0 log( cosh(T/T0) ) .
Notice that V only approaches V0 asymptotically, it never really gets there.
For T = T0, you get V is about 3/4 the terminal velocity and you will have

LeeH

(published on 03/27/2010)

## Follow-Up #1: time to reach terminal velocity

Q:
Hello, I have a question for you regarding drag forces. When an object (such as a raindrop) falls from a certain height, let's say y=y1, a drag force is act on the raindrop (along with gravitational forces) so that there is a special time where the drag force has actually equaled its magnitude with the weight magnitude, making the net forces that act on the body equal to zero, and from now on the velocity of the raindrop is constant (terminal velocity). However, it is clear that not all objects can achieve a termnial, constant velocity if there is not enough time for it to occur. Therefore, i'd like to know how can we find that specific time when the net forces are equal to zero (suppose we have all others parameters such as mass, density, etc). Thanks in advance, Yoav
- Yoav Elizur (age 22)
Tel-Aviv, Israel
A:

In the standard version of that problem, where the air density is pretty much constant over the range of the fall, the velocity never quite reaches the terminal velocity. It creeps up closer and closer, with the terminal velocity just being the limit. The time it takes to get very close to the terminal velocity VF is a few times VF/g, where g is the gravitational acceleration. Lee discusses that above. The actual form for the time-dependent velocity v(t) is v(t)+VF*tanh(gt/VF). See  for more discussion.

When something falls from very high up, where the air is less dense, the terminal velocity starts off high and goes down as the object falls into denser air. Then at some point the acceleration does reach zero, as the object ceases to accelerate and starts to decelerate.  (see )

Mike W.

(published on 12/18/2013)