# Parachutist's descent with air resistance

I have partially solved the following problem, taken from this book:

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?

My partial solution is that, for a fixed value of $\eta$, we have:

$$A \dot x = \tanh(C - Bt) \\ \eta x = \ln \frac {e^{C - Bt}} {1 + e^{2C - 2Bt}} + D$$

where $A = \sqrt {\eta / g}$, $B = \sqrt {\eta g}$, and $C$ and $D$ are constants of integration.

Suppose that I have computed constants $A_1, B_1, C_1, D_1$ for a given $\eta_1 > 0$ and initial conditions $x_1(0) = x_0$ and $\dot x_1(0) = v_0$. I shall compute constants $A_2, B_2, C_2, D_2$ for a given $\eta_2 > \eta_1$ and initial conditions $x_2(t_0) = x_1(t_0)$ and $\dot x_2(t_0) = \dot x_1(t_0)$. Expanding the last condition, we get

$$A_2 \tanh (C_1 - B_1 t_0) = A_1 \tanh (C_2 - B_2 t_0)$$

Since $\eta_2 > \eta_1$, it follows that $A_2 > A_1$. If $t_0$ is large enough, the absolute value of $\tanh (C_1 - B_1 t_0)$ will be close enough to $1$ to make absolute value of $\tanh (C_2 - B_1 t_0)$ greater than $1$, which is a contradiction. What should I conclude from this?

• The limit of the speed can be found by setting $\ddot{x}$ to zero to get $v_\max = \sqrt{g \over \eta}$. – copper.hat Mar 29 '18 at 16:49
• @copper.hat I have indeed found that limit by myself. What I'm having trouble with “matching” the two solutions with different values of $\eta$ before and after $t = t_0$. – pyon Mar 29 '18 at 16:52

## 2 Answers

Your initial velocity for the second phase of the fall can be greater than the terminal velocity. This means the sign is swapped in the integration and the solution isn't in fact a hyperbolic tan.

• Ah, nice, thanks! – pyon Mar 29 '18 at 17:07

The real problem is that I made the unwarranted assumption that $A$ has to be greater than 0. Here is the right way to do it: http://mathb.in/23850.