# 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

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.