[Physics] Equation for a falling body including terminal velocity

Question

I'm making an app that times how long it takes a pebble to fall, then calculates the distance it fell.

I noticed that the simple $f(t) = \frac{1}{2}gt^2$ was becoming increasingly inaccurate as the distance got higher, so I'm curious if there's a standard formula accounting for terminal velocity.

What I've come up with so far

The wikipedia page on terminal velocity lists the formula for terminal velocity ($V_t$) as:

$$
V_t = \sqrt{\frac{2mg}{\rho AC_d}}
$$

I'm approximating that the pebble weighs $m=5g$ with a projected area $A=2cm^2$. $C_d=0.47$ for spheres and $\rho =1.204$ at $20^{\circ}C$, so that gives us:

$$
V_t = \sqrt{\frac{2\cdot0.005\cdot9.81}{1.204\cdot 0.0004\cdot 0.47}} \approx 20.82 m/s
$$

Seems pretty above board so far. But now I need to combine $f(t) = \frac{1}{2}gt^2$ with $V_t = 20.82 m/s$ so $f(t)$ "grows asymptotically" to $V_t$. I don't know how to do that, but playing around with my computer's graphing program got me this:

$$
f(t) = \frac{(v_t-\frac{1}{5})\cdot x^2+1}{x^2+5}-\frac{1}{5}
$$

($\frac{1}{2}gt^2$ is green, $V_t$ is dashed, my made up formula is blue)

My approximated formula seems… close? I could take some measurements and validate this new formula experimentally, but I can't imagine I'm the first person to need to approximate distance given time for a falling object.

Also: this graph made it pretty clear that after ~2 seconds of the pebble falling $\frac{1}{2}gt^2$ (the green line) starts getting grossly inaccurate.

tl;dr: what's the formula for the velocity of a falling object with respect to time given $g, \rho, A$, and $C_d$?

Answer 1

First, let's clarify the origin of the different expressions of terminal velocity and velocity as a function of time for a falling body.

The friction force is expected to be an increasing function of the body velocity, and as a result, there exists a velocity for which this force balances exactly the gravity $mg$. Now, in order to calculate this terminal velocity, or to recover $t\mapsto v(t)$, it is necessary to know more about the expression of the friction force. The most common modellings of this force are :

• The quadratic law, $F = \frac{1}{2}AC_d \rho v^2$. Those are the same variables as in your question. Since $C_d$ is roughly constant in the high velocity regime, the $v^2$ factor approximately accounts for the whole velocity dependence, and the force is proportional to $v^2$. We can also write $F = K_{square}v^2$. But then, this applies to fast enough motions, those for which the Reynolds number $Re$ is large. Since it is proportional to the inverse of the viscosity of the medium, and as the atmosphere air is not very viscous/dense, $Re$ is effectively large for most fall-in-the-air experiments and this is the right expression for $F$
• The linear law, $F = K_{linear}v$. This laws applies to the low velocity regime, and is mostly relevant to high viscosity fluids.

While the first expression indeed yields the same expression as you employed for the terminal velocity (the square root being originated from the square power on $v$ in $F(v)$), it is not consistent with an exponential time dependence. The correct expression is, as given by wikipedia : $$v(t) = \sqrt{ \frac{2mg}{\rho A C_d} } \tanh \left(t \sqrt{\frac{g \rho C_d A}{2 m}} \right)$$ The exponential time evolution of the velocity would be obtained using the linear expression for the force (then the equation of motion is a 1st order linear differential equation in $t\to v(t)$).

You can integrate the correct expression for $v(t)$ to recover the fall distance with time (it will of the form $d(t) \propto \ln (\cosh (t/\tau))$)

Regarding your comparison in http://i.stack.imgur.com/j7Uhh.png, it does not make sense, since you are comparing a velocity (which tends to a constant) to a distance ($\frac{1}{2}gt^2$).