Velocity of Freefall Body

Direction : Derive a differential equation that tells you the velocity of a falling body at any given time. Two forces act on the body. Gravity pulls it down, and air resistance pushes back against its motion.

(Assume the condition where you should not ignore the air resistance)

What is the governing rule ?

The governing rule here is Newton's second law, written in words below. The work is to list every force on the falling body and to express the acceleration through the velocity, because the velocity is the unknown we want.

Total Force applied to a body = Motion of the body

The diagram below builds the equation step by step. On the left, it lists the two forces: the pulling force of gravity, mg, and the air resistance, -kv. Their sum is mg - kv. On the right, it replaces the acceleration a with dv/dt, so ma becomes m dv/dt. The note at the bottom explains the minus sign. The air resistance acts against gravity, and gravity is taken as the positive direction.

Forces on a falling body with a parachute: gravity mg and air resistance -kv, and ma written as m dv/dt

Setting the total force equal to m dv/dt gives the governing equation. The equation chain below writes it in three steps, from F = ma to m dv/dt = mg - kv.

F = ma leads to m dv/dt = mg - kv

Note : This can be one of the simplest modeling example, but it doesn't mean that you should be able to understand everything here in 5 minutes and you are dumb if you don't understand this so quickly. I hope you would have a lot of questions that has not clearly been explained in my explanation above and that has not been treated in your math course.

For example, my own questions are like this.

The explanation said the air resistance can be expressed as -kv, but how did you know that ? Jugding from the expression, it seems that the air resistance is determined only by the velocity ? How did you guarantee that the resistance is not influenced by other factors like acceleration, mass of the falling body ? How about the shape of the falling body or paraschute ? Doesn't this influence the resistance force ? Judging from our common sense, the shape/size of the paraschute definately influence the resistance force. How come this term is not shown in the mathematical expression, -kv ? what does 'k' mean ? How can we determine this value ? what kind of factors are involved in determining k value (Actually, the shape of paraschute or falling body can be a very important factors to determine the k value).

Like this, we can have an endless list of questions boggling in your mind if you are really doing the mathematical modeling in stead of rotely memorizing whatever your textbook or teach tells.

I am not saying that you always have to raise your hand in your math class and argue with your math teacher until you get the answers to all of these kind of questions. These are the questions you continue to investigate to expand this kind of basic model to a better/real life models.

Gravity is taken as the positive direction : the velocity v is positive while the body falls, so the air resistance enters as -kv.
The unknown is the velocity, not the position : writing a as dv/dt gives a first order equation in v.
The air resistance -kv is an assumption : the questions in the note above all come from this one choice.

How does the velocity change over time ?

The governing equation tells us how fast the velocity changes, but not the velocity itself. Let's solve it for a body that starts from rest, and then read the terminal velocity from the result.

The equation m dv/dt = mg - kv is first order and linear, and its variables separate. Divide both sides by mg - kv and integrate. With v(0) = 0, the solution is v(t) = (mg/k)(1 - e^-kt/m).

As t grows, the exponential term dies out, and the velocity approaches v_T = mg/k. This is the terminal velocity. At that speed, the air resistance kv equals the weight mg. So the net force is zero, and the body stops accelerating. The time constant is m/k. After one time constant, the velocity reaches 63 percent of v_T, and after three time constants it reaches 95 percent.

For example, take a body with its parachute open, with m = 80 kg, k = 160 kg/s and g = 9.8 m/s². The terminal velocity is 80 x 9.8 / 160 = 4.9 m/s, and the time constant is 80/160 = 0.5 s. So the body settles at about 4.9 m/s within about 1.5 s. Without air resistance, the same body would already fall at 14.7 m/s after 1.5 s, and its speed would keep growing as gt.

The terminal velocity is where the two forces balance : kv_T = mg, so v_T = mg/k.
A larger k lowers the terminal velocity and shortens the settling time : both the terminal velocity mg/k and the time constant m/k shrink as k grows.
The solution contains free fall as a limit : for a very small k, the exponential expands to 1 - kt/m, and the solution becomes v = gt.

What changes if the air resistance is not -kv ?

The note above asks where -kv comes from and what k depends on. Let's give a short answer, because it also shows how a basic model grows into a better one.

The linear form -kv fits slow motion through a viscous fluid, such as a small particle or a droplet falling in air. For a person or a parachute, the air resistance grows closer to the square of the speed. A common model writes its size as (1/2)ρC_dAv². Here ρ is the air density, A is the area facing the flow and C_d is the drag coefficient, which depends on the shape. So the shape and the size of the parachute enter through C_d and A. The mass and the acceleration do not appear in the resistance itself. The mass enters only through the ma side of Newton's law.

With c = (1/2)ρC_dA, the governing equation becomes m dv/dt = mg - cv². It is no longer linear, but its variables still separate. Starting from rest, the solution is v(t) = v_T tanh(gt/v_T), with the terminal velocity v_T = √(mg/c).

The two models agree that a terminal velocity exists, but they approach it at different rates. Take the same terminal velocity of 4.9 m/s for both. The linear model needs about 1.15 s to reach 90 percent of it, and the quadratic model needs about 0.74 s.

k is not a constant of nature : it collects the shape, the size and the fluid, which is where the parachute enters the model.
The form of the resistance is a modeling choice : -kv suits slow, small bodies, and -cv² suits fast, large ones.
Both models balance at a terminal velocity : the resistance grows until it equals the weight mg, whatever its form.