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 $t0 > 0$. Model this situation by assuming $\eta = \eta1$ for $0 < t < t0$ and $\eta = \eta2 > \eta1$ for $t > t0$ and match the solutions at $t = t_0$. What does the solution look like?

Solution: Introducing the helper variable $v = \dot x$, we have

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

Rewriting the constants in terms of $A = \sqrt{\eta / g}$ and $B = \sqrt{\eta g}$, we have

$$\frac {A\ dv} {1 - A^2 v^2} = -B\ dt \ \tanh^{-1}(Av) = C - Bt \ Av = \tanh(C - Bt)$$

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 $v_\infty = -1/A = \sqrt{g / \eta}$, because

$$Av\infty = \lim{t \to \infty} Av(t) = \lim{t \to \infty} \tanh(v0 - bt) = -1$$

Let us finish solving the differential equation. We have

$$A\ dx = \tanh (C - Bt)\ dt$$

Introducing the helper variable $s = C - Bt$, we have

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

Introducing the helper variable $r = e^s$, 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

$$\eta x - D = \ln \frac r {1 + r^2} = \ln \frac {e^s} {1 + e^{2s}} = \ln \frac {e^{C - Bt}} {1 + e^{2C - 2Bt}}$$

Thursday, 29 March 2018 16:18 GMT