How long does a skydiver take to reach terminal velocity?

For my 16th birthday I went indoor skydiving in a vertical wind tunnel. The instructor explained that the air in the tunnel was moving at about 200 km/h, “the same as your terminal velocity in a real jump”, and that in a real skydive “it takes about ten seconds to get there”. I have a tandem jump booked for next summer, and I wanted to know whether that ten seconds is right — and what “get there” even means, because in the models we met in class, a falling object only approaches its terminal velocity and never actually reaches it.

The aim of this exploration is to model the speed of a skydiver in free fall with a differential equation, find how long it takes to reach 95% (and 99%) of terminal velocity, and test how much the answer depends on the assumptions in the model.

I chose 95% because it is a common way of defining when an exponential-type process is “nearly finished”, but I will check whether the choice matters. I expected air resistance to be proportional to the square of the speed, because that is what my physics teacher said applies to large objects moving fast, but I will compare it with the simpler linear model to see how much difference it makes.

I modelled the skydiver as a particle of mass \(m\) falling vertically from rest. I define:

\(t\) = time since leaving the aircraft (s), \(t \ge 0\);
\(v\) = downward speed (m s⁻¹), \(v \ge 0\);
\(g = 9.81\) m s⁻² (gravitational field strength, assumed constant);
\(k\) = drag constant (kg m⁻¹), assumed constant for now.

Two forces act: the weight \(mg\) downwards and air resistance \(kv^2\) upwards. By Newton's second law,

\[ m\frac{dv}{dt} = mg - kv^2 . \qquad (1) \]

When the forces balance the acceleration is zero, so the terminal velocity \(v_T\) satisfies \(mg = kv_T^2\), giving \(v_T = \sqrt{mg/k}\). This lets me rewrite (1) in terms of \(v_T\), which is the quantity I actually know:

\[ \frac{dv}{dt} = g\left(1 - \frac{v^2}{v_T^2}\right). \qquad (2) \]

For a skydiver in the belly-down position, 55 m s⁻¹ (about 200 km/h, as my instructor said) is a typical terminal velocity, so I used \(v_T = 55\) m s⁻¹ and \(m = 80\) kg, which gives \(k = \frac{mg}{v_T^2} = \frac{80 \times 9.81}{55^2} \approx 0.259\) kg m⁻¹. Equation (2) shows that the answer to my question depends only on \(v_T\) and \(g\), not on the mass separately — which surprised me, and which means my result applies to anyone with the same terminal velocity.

Equation (2) is separable. Separating the variables and integrating from the start of the jump (\(v = 0\) at \(t = 0\)):

\[ \int_0^{v} \frac{v_T^2}{v_T^2 - u^2}\,du = \int_0^{t} g\,ds . \]

To integrate the left-hand side I used partial fractions. Since \(v_T^2 - u^2 = (v_T - u)(v_T + u)\),

\[ \frac{v_T^2}{(v_T - u)(v_T + u)} = \frac{A}{v_T - u} + \frac{B}{v_T + u} \;\Rightarrow\; v_T^2 = A(v_T + u) + B(v_T - u). \]

Substituting \(u = v_T\) gives \(A = \frac{v_T}{2}\), and \(u = -v_T\) gives \(B = \frac{v_T}{2}\). Therefore

\[ \frac{v_T}{2}\Big[-\ln(v_T - u) + \ln(v_T + u)\Big]_0^{v} = gt \;\Rightarrow\; \frac{v_T}{2}\ln\frac{v_T + v}{v_T - v} = gt , \]

where the logarithms are defined because \(0 \le v < v_T\) throughout the fall (the speed starts below \(v_T\) and, by (2), can never cross it, since \(dv/dt = 0\) at \(v = v_T\)). Rearranging,

\[ \frac{v_T + v}{v_T - v} = e^{2gt/v_T} \;\Rightarrow\; v = v_T\,\frac{e^{2gt/v_T} - 1}{e^{2gt/v_T} + 1} = v_T\tanh\left(\frac{gt}{v_T}\right). \qquad (3) \]

The last step uses the definition \(\tanh x = \frac{e^x - e^{-x}}{e^x + e^{-x}}\), which I found by multiplying the numerator and denominator by \(e^{-gt/v_T}\). I had not met \(\tanh\) before; its graph rises from 0 and levels off at 1, which is exactly the behaviour I expected for the speed. I checked (3) by differentiating it: \(\frac{dv}{dt} = g\,\text{sech}^2\frac{gt}{v_T} = g\left(1 - \tanh^2\frac{gt}{v_T}\right) = g\left(1 - \frac{v^2}{v_T^2}\right)\), which is equation (2).

As \(t \to \infty\), \(\tanh(gt/v_T) \to 1\), so \(v \to v_T\) but \(v < v_T\) for every finite \(t\). This confirms mathematically that terminal velocity is never actually reached — so the instructor's “ten seconds” must mean something like “close enough”.

With quadratic air resistance, an 80 kg skydiver with a terminal velocity of 55 m s⁻¹ reaches 95% of that speed about 10.3 s after leaving the aircraft, having fallen about 359 m, and 99% after about 14.8 s. Because \(v = v_T\tanh(gt/v_T)\) is always below \(v_T\), “reaching terminal velocity” has to be defined by a threshold, and my answer depends on that choice: the instructor's “ten seconds” matches the 95% definition.

The biggest assumption is the form of the drag. Linear drag would have given 16.8 s — very different — but my shuttlecock experiment supports the quadratic model, and the model's predictions agree with skydivers' rules of thumb. Assuming constant air density changes the total free-fall time by about 3 s but barely affects the 95% time. The weakest link is the value of \(v_T\): it varies with body position and clothing, and because \(t_{95}\) is proportional to \(v_T\), a 10% error in \(v_T\) gives a 10% error in my answer.

What I learned most is how much a model's answer depends on a sensible definition — the question “when do you reach terminal velocity?” has no answer until you decide what “reach” means. Next summer I plan to wear an audible altimeter with a data log, and I would like to compare my own fall with this model.