I'm trying to figure out how high an object would have to be dropped to reach it's terminal velocity. Specifically if an object had a terminal velocity of 520 mph, how would I figure out how high it would have to be dropped to reach that speed. I know there's probably more factors that go into this then just knowing it's terminal velocity, but if someone could just help me out with an equation and what to input that would be great and I can handle it from there. I've searched every where for an equation that would solve it, and for some reason I can't seem to picture the process in my head to figure it out.

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    $\begingroup$ Assume high Reynolds number approximation for drag $\propto -v|v|$, solve differential equation for velocity. Discover that body never actually reaches its terminal velocity. So now we ask: how close to its terminal velocity counts as reaching it? $\endgroup$ Jul 21 '15 at 20:59
  • $\begingroup$ Lets just say were solving for any specific speed. Not the terminal velocity. I'm still not sure what to do. I'm not really into physics, but I really want to figure this out. $\endgroup$ Jul 21 '15 at 21:29

Okay, so here's the basic physics: it comes down to differential equations.

Big objects in air tend to see $v^2$ drag. Plugging this into $F = m a$, this means that something which is falling straight downward with speed $v$ will obey the nonlinear differential equation: $$\frac {dv}{dt} = \frac 1\lambda v^2 - g$$for some length $\lambda$.

The terminal velocity is the one where $dv/dt = 0$ hence $v_t^2 = \lambda ~ g,$ giving a simple way to compute $\lambda$ if you know the terminal velocity and the graviational acceleration

Here's the part that you're not going to like: actually solving this equation gives$$\int ~\frac{dv}{v_t}~ \frac{-1}{1 - (v/v_t)^2} = \frac {v_t} \lambda ~ t + C~.$$Then, choosing $v = v_t~\tanh{u}$ we find$$-u = \frac t\tau + C; ~~~v = -v_t~\tanh~\frac{t - t_0}{\tau}~.$$The reason this is frustrating is not that the hyperbolic tangent $\tanh$ is an awful function or anything -- it's actually very easy to work with and quite well-behaved! -- but because it means that we never actually attain the terminal velocity, we just get closer and closer to it. So the answer is trivially $\infty$.

With that said, if you want to find, say $v = 0.9 ~ v_t$ many calculators support the $\operatorname{atanh}$ operation that inverts the hyperbolic tangent; for example Wolfram Alpha gives $1.47222\dots$. With this time constant $\tau = \lambda / v_t$ you know that from rest, it takes $1.47~\tau$ to achieve that speed.

As for integrating once more to achieve a vertical position, it is not too hard to do that integral because $\tanh = \sinh / \cosh$ so you can simply use $u = \cosh~\frac{t - t_0}\tau$ for the substitution. So that gives a simple result of $$y(t) = y_0 - \lambda ~ \ln\left(\cosh~\frac{t - t_0}\tau \right).$$Hope that helps. Nonlinear differential equations rapidly get quite hairy; this is one of the few that is both really straightforward and which offers some really useful/invertible/manageable answers.


Lets just say were solving for any specific speed. Not the terminal velocity. I'm still not sure what to do.

It might help to start with the speed an object attains after falling a short distance, ignoring air resistance and assuming the effect of gravity is constant. If we fall from Mount Everest (about 8000 meters) to sea level, we can treat gravity as being close to constant.

For this distance, without air friction, we can say that $$v=\sqrt{2gh}$$ where $g$=9.8 meters per second squared, and $h$ is the distance falling in meters. That gives you $v$ in meters per second. A quick calculation to get to mph is to multiply that number by 2.2.

Example: A baseball falling from rest for a distance of 400 m (about 1/4 mile) would reach a speed of $$\sqrt{2(9.8)(400)} = 88.5\text{ m/s } \simeq 195\text { mph}.$$

Considering that air resistance on a baseball is significant at half that speed, ignoring it isn't a good approximation for that high a drop. But it does tell you the minimum height you would need to even have a chance of reaching a certain speed, and it tells you that you probably won't reach the speed this formula calculates.

I'll let others chime in on more specific air resistance effects and calculations.


For a friction less drop we simply use Newtons law:

$$\frac{dv}{dt} = g$$

and solving for $v$ gives $v(t) = gt$.

Introducing the friction component into the differential equation, I make the approximation that the drag is laminar:

$$ \frac{dv}{dt} = g - \frac{\gamma}{m} v$$

The solution of this is: $v(t) = \frac{mg}{\gamma} - \frac{mg}{\gamma} e^{-\frac{\gamma}{m} t} $, assuming that $v(0) =0$.

As $t$ goes to infinity $v \rightarrow \frac{mg}{\gamma}=:c$ where $c$ is the terminal velocity.

We rewrite $v(t) = c - c e^{-\frac{g}{c} t} $

If we solve for the hight $h$ we get: $h(t) = -ct - \frac{c^2}{g} e^{-\frac{g}{c} t} + h_0 + \frac{c^2}{g}$.

Now we want to approximate the hight such that the terminal velocity gets reached, say with an accuracy of $\epsilon$. First we will use the equation for the speed to determine the time $\tau$ for the fall we find at:

$$ \tau = -\frac{c}{g} \log(\frac{\epsilon}{c}) $$

Substituting this into the equation for position $0 = h(\tau)$, we arrive at an expression for $h_0$:

$$h_0 = \frac{c^2}{g} \big( \frac{\epsilon}{c}- \log(\frac{\epsilon}{c}) -1 \big)$$

I left many sub-steps out as you are probably not that interested.

So for $c = 520\ \text{mph} \approx 231\frac{m}{s}$ and $g\approx 1\frac{m}{s^2}$ and the accuracy $\epsilon = 1\frac{m}{s}$, thus the speed reached would be $230\frac{m}{s}$, the result is:

$$ h \approx 24\ \text{km} = 15\ \text{miles}$$

  • $\begingroup$ It is unlikely that they are considering the terminal velocity of a bacterium in water, hence $\gamma~v/m$ is not the applicable drag force. $\endgroup$
    – CR Drost
    Jul 21 '15 at 23:06
  • $\begingroup$ I know that proportianal to $v$ is an for laminar drag, the results will still give a good approximation. @ChrisDrost this is no reason to down vote. $\endgroup$
    – john
    Jul 22 '15 at 12:14
  • $\begingroup$ I mean, you do have an approximately right result for $\epsilon/c \approx 1$, although it has a linear deviation from that because it does much worse for that regime than the free fall equation $c^2/g (1 - \epsilon/c)^2$. By the time you get to $\epsilon/c \approx 0$ your heights are about 200% of the correct ones. So you're saying 24 km for a problem where the correct answer is only 13km. It's good for a rough order of magnitude I guess? $\endgroup$
    – CR Drost
    Jul 22 '15 at 13:42

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