**Problem:** Consider the free fall with air resistance modeled by

$$\ddot x = \eta \dot x^2 - g$$

Solve this equation. (Hint: Introduce the velocity $v = \dot x$ as the new dependent variable.) Is there a limit to the speed the object can attain. If yes, find it. Consider the case of a parachutist. Suppose the chute is opened at a certain time $t*0 > 0$. Model this situation by assuming $\eta = \eta*1$ for $0 < t < t*0$ and $\eta = \eta*2 > \eta*1$ for $t > t*0$ and match the solutions at $t = t_0$. What does the solution look like?

**Solution:** Let $v = \dot x$. Substituting in the original equation, we have

$$\frac {dv} {\eta v^2 - g} = dt \ \frac {dv} {1 - \frac \eta g v^2} = -g\ dt$$

Let $a \in \mathbb R$ such that $a^2 = \eta / g$. Substituting in the previous equation, we have

$$\frac {a\ dv} {1 - a^2 v^2} = -ag\ dt \ \tanh^{-1}(av) = b - agt \ av = \tanh(b - agt)$$

where $b$ is a constant of integration.

Using what we have calculated so far, we can establish that, in fact, there is a limit to the speed the object can attain. This limit is $v_\infty = -1/|a| = -\sqrt{\eta / g}$, because

$$av*\infty = \lim*{t \to \infty} av(t) = \lim_{t \to \infty} \tanh(b - agt) = -\text{sgn }a$$

Let us finish solving the differential equation. We have

$$a\ dx = \tanh (b - agt)\ dt$$

Let $s = b - agt$. Substituting in the previous equation, we have

$$\eta\ dx = a^2 g\ dx = -\tanh s\ ds = \tanh(-s)\ ds = \frac {1 - e^{2s}} {1 + e^{2s}} ds$$

Let $r = e^s$. Substituting in the previous equation, we have

$$\eta\ dx = \frac {1 - r^2} {1 + r^2} ds = \frac {1 - r^2} {r (1 + r^2)} dr = \left[ \frac 1 r - \frac {2r} {1 + r^2} \right] dr$$

Integrating and systematically reverting all variable substitutions, we have

$$\eta x - c = \ln \frac r {1 + r^2} = \ln \frac {e^s} {1 + e^{2s}} = \ln \frac {e^{b - agt}} {1 + e^{2b - 2agt}}$$

where $c$ is another constant of integration.