Plane Crash Problem: How Long Does It Take to Fall?
How fast does a falling object end up moving, and how long does the fall take? This question came up in a conversation in Shenzhen: does a falling plane reach a higher terminal velocity than a falling person? We first solve the problem with constant air density, then let the density decrease exponentially with altitude.
Constant air density
Take downwards as positive and model air resistance as quadratic drag, \(kv^2\) with \(k > 0\). Newton’s second law gives a separable equation:
\[\begin{aligned} m \frac{dv}{dt}&=mg-kv^2 \\ \frac{m\,dv}{mg-kv^2}&=dt \\ t&=\int_{0}^{v} \frac{m\,dv'}{mg-kv'^2}, \end{aligned}\]where we assumed the object starts from rest, \(v(0)=0\).
To evaluate the integral, substitute \(u=\sqrt{\frac{k}{mg}}\, v\), so that \(dv = \sqrt{\frac{mg}{k}}\, du\):
\[t = \frac{m}{mg}\sqrt{\frac{mg}{k}} \int_0^{u} \frac{d u'}{1-u'^2} = \sqrt{\frac{m}{gk}}\, \tanh^{-1}\!\left(v \sqrt{\frac{k}{m g}}\right).\]Inverting,
\[v(t)=\sqrt {\frac{mg}{k}}\,\tanh \left( t \sqrt{\frac{gk}{m}} \right).\]As \(t \to \infty\), \(\tanh \to 1\), so the terminal velocity is
\[v_\text{T}=\sqrt{\frac{mg}{k}}.\]To find \(k\), use the drag equation for turbulent flow, \(F_{\mathrm{d}}=\frac{1}{2} \rho\, c_{\mathrm{d}} A v^2\), so \(k = \frac12 \rho\, c_d A\) and
\[v_\text{T}=\sqrt{\frac{2mg}{\rho\, c_d A}}.\]The terminal velocity grows like \(\sqrt{m/(c_d A)}\). A plane’s mass per unit frontal area is far larger than a person’s, and the differences in drag coefficient and area cannot compensate, so a falling plane reaches a much higher terminal velocity.
Altitude-dependent air density
Air thins out with height. A simple model is an exponential atmosphere,
\[\rho(h)=\rho_0\,e^{-h/H},\]where \(H\) is the scale height (about \(8\text{ km}\) for Earth). We now look for the speed \(v\) as a function of the altitude \(h\).
With downwards still positive, the equations of motion are
\[m\frac{dv}{dt}=m g - k\,e^{-h/H}\,v^2, \qquad \frac{dh}{dt}=-v, \qquad k=\tfrac12\,\rho_0\,c_d\,A.\]1. Change of independent variable
Trade the time derivative for an \(h\)-derivative:
\[\frac{dv}{dt} =\frac{dv}{dh}\,\frac{dh}{dt} =-\,v\,\frac{dv}{dh}.\]Setting \(u(h)=v(h)^2\), the equation of motion becomes linear in \(u\):
\[\frac{du}{dh} -\frac{2k}{m}\,e^{-h/H}\,u =-2g.\]2. Integrating factor
The integrating factor is
\[\mu(h) =\exp\left(-\frac{2k}{m}\int e^{-h/H}\,dh\right) =\exp\left(\frac{2kH}{m}\,e^{-h/H}\right).\]Multiplying the equation by \(\mu\) and integrating from the initial altitude \(h_0\), where the object starts from rest (\(u(h_0)=0\)),
\[\frac{d}{dh}\bigl(u\,\mu\bigr) =-2g\,\mu(h) \quad\Longrightarrow\quad u(h)\,\mu(h) =-2g \int_{h_0}^{h}\mu(s)\,ds.\]The remaining integral is an exponential integral. With \(w = \frac{2kH}{m}e^{-s/H}\) we have \(ds = -H\,dw/w\), so
\[\int e^{\frac{2kH}{m}e^{-s/H}}\,ds = -H\,\operatorname{Ei}\left(\frac{2kH}{m}e^{-s/H}\right) + C.\]3. Speed as a function of altitude
Define
\[y = \frac{2kH}{m}\,e^{-h/H}, \qquad y_0 = \frac{2kH}{m}\,e^{-h_0/H}.\]Then
\[v(h)^2 = 2gH\, e^{-y}\left[\operatorname{Ei}(y) - \operatorname{Ei}(y_0)\right], \qquad \boxed{ v(h) =\sqrt{2gH}\; e^{-y/2}\, \sqrt{ \operatorname{Ei}(y) -\operatorname{Ei}(y_0) } }.\]Since \(h < h_0\) we have \(y > y_0\), and \(\operatorname{Ei}\) is increasing for positive arguments, so the square root is real.
4. Time as a function of altitude
The time to fall from \(h_0\) to \(h\) is
\[t(h)=\int_{h}^{h_0}\frac{dh'}{v(h')},\]and unlike the constant-density case this integral has no elementary closed form, so in practice it is evaluated numerically. For example, a skydiver with \(m = 80\text{ kg}\) and \(k = 0.25\text{ kg m}^{-1}\), falling from rest at \(h_0 = 10\text{ km}\) with \(H = 8\text{ km}\), is moving at \(64.3\text{ m s}^{-1}\) when passing \(2\text{ km}\), about \(104\text{ s}\) later. A direct numerical integration of the equations of motion agrees with both numbers.
5. Recovering the constant-density result
As \(H\to\infty\) the atmosphere becomes uniform. Both \(y\) and \(y_0\) then tend to infinity, while their difference stays finite:
\[y - y_0 = \frac{2kH}{m}\left(e^{-h/H}-e^{-h_0/H}\right) \longrightarrow \frac{2k}{m}(h_0 - h).\]For large arguments \(\operatorname{Ei}(x) \sim e^x/x\), so
\[v(h)^2 \approx 2gH\left[\frac{1}{y} - \frac{e^{-(y-y_0)}}{y_0}\right] \longrightarrow \frac{mg}{k}\left[1 - e^{-2k(h_0-h)/m}\right].\]This is exactly the constant-density result written in terms of the distance fallen, \(s = h_0 - h\). Indeed, integrating \(v(t)=\sqrt{mg/k}\,\tanh\bigl(t\sqrt{gk/m}\bigr)\) gives \(s(t) = \frac{m}{k}\ln\cosh\bigl(t\sqrt{gk/m}\bigr)\), so \(e^{-2ks/m} = \operatorname{sech}^2\bigl(t\sqrt{gk/m}\bigr)\) and \(v^2 = \frac{mg}{k}\left(1-e^{-2ks/m}\right)\). The exponential-atmosphere solution therefore reduces smoothly to the familiar \(\tanh\) law.
Exercises
- Why is the constant-density equation separable? (Undergraduate, year 1)
- Repeat the analysis for air resistance proportional to \(v\). (Undergraduate, years 1–2)
- With no air resistance at all, what are the “terminal velocities” of the plane and the person? Why? (Secondary school)
Changes on 25 October 2026
What was wrong. The boxed formula for the fall time when the air density decreases with height was wrong. For a skydiver (80 kg) falling from 10 km to 2 km it gives about \(-3 \times 10^{-6}\) s, a negative time, while integrating the equations of motion gives about 104 s. The formula for the speed was correct.
What changed. The time formula is replaced by the correct integral, and the constant-density derivation is cleaned up. The difficulty label “HKDSE F4” is now “secondary school”.