What is the terminal speed of a falling object subject to drag proportional to its velocity if it travels a certain distance in a given time? This is problem 7.52 from the first volume of Alonso and Finn's Fundamental University Physics.
The given data is that a body falls 108 meters in five seconds. We are to find the terminal velocity if it is dropped from rest and feels a drag force proportional to its velocity.
Work so far:
Finding an analytical solution for this is straightforward. The equation of motion is
$$ma = mg - bv$$
where $b$ is the proportionality constant.
The acceleration is
$$a = g - \frac{b v}{m}$$
and we can write
$$\begin{align}dv &= \left(g - \frac{b v}{m}\right) dt. \\
dv &= -\frac{b}{m} \left(v - \frac{m g}{b}\right) dt. \end{align}$$
We can integrate to get an expression for the velocity.
$$\begin{align} \int_0^v \frac{dv'}{v' - \frac{m g}{b}} &= \int_0^t - \frac{b}{m} dt'. \\
\ln \left(\frac{m g - b v}{m g}\right) &= - \frac{b}{m} t. \\
m g - b v &= m g e^{- \frac{b}{m} t}. \\
v &= \frac{m g}{b} \left(1 - e^{- \frac{b}{m} t}\right). \end{align}$$
Analytically, clearly the terminal speed is $v_\mathrm{T} = \frac{m g}{b}$. I assume that since information is given about distance traveled and the time elapsed, one is meant to find a numerical answer. The above can be integrated again to find the position. Assuming I've gotten that part right, we have
$$\begin{align} x &= \frac{m g}{b} t - \frac{m}{b} \frac{mg}{b} \left(1 - e^{-\frac{b}{m} t} \right). \end{align}$$
Obviously the expression for $v$ itself appears in this expression and that is how I feel I should be able to proceed to somehow solve for $\frac{m}{b}$ or something, but I just can't see where to go from here. I can't come up with any further assumptions since I don't know anything about the body itself or how $b$ depends on its properties. I'm greatly thankful for any help on what I'm overlooking. I've been working through this text from the beginning, and I have realized it does have some pretty serious issues with errors in the odd-number solutions at the back (like the same errors reappearing frequently, for instance regarding tetrahedron geometry with questions in both chapters 3 and 4) and a few questions that seem unclear about what they are actually asking. So this may just be poorly constructed, as bold as it may sound to say so.
 A: Interesting problem! I show here that you do not have to solve for $b$ and $m$, but for their ratio $k = b/m$ which is unique for the given set of data.
Let me just rewrite your equations
$$
\begin{aligned}
x(t) &= \frac{1}{k} g t - \frac{1}{k^2} g \bigl( 1 - e^{-kt} \bigr) \\
v(t) &= \frac{1}{k} g \bigl( 1 - e^{-kt} \bigr)
\end{aligned}
$$
You have correctly identified that velocity appears in the displacement
$$x(t) = \frac{1}{k} \bigl( g t - v(t) \bigr)$$
and the terminal velocity is
$$\boxed{v_f = g t_f - k x_f}$$
where $x_f = x(t_f)$ and $v_f = v(t_f)$. The only problem here is that value of $k$ is unknown. I am not sure if this is possible to solve analytically, but it is possible to show that there is only one solution for given $x_f$ and $t_f$
$$\underbrace{(x_f/g) k^2 - t_f k + 1}_{f_1(k)} = \underbrace{e^{-t_f k}}_{f_2(k)}$$
In the above equation the only unknown is $k$ which is determined from the intercept between quadratic function $f_1(k)$ and exponential function $f_2(k)$. There are two solutions for the above equation: (i) trivial solution is $k = 0$, and (ii) the other solution is what you need to calculate $v_f$. Bottom line is that there is only one non-trivial solution $k^\star$ to the above equation which means final velocity solution $v_f$ is unique for given $x_f$ and $t_f$.
Figure below shows how to determine unknown parameter $k$ by simple numeric procedure, which you could also do by hand in few (or dozen) iterations. From this we get $k^\star = 0.0781$ and the terminal velocity is $v_f = 40.57 \text{ m/s}$.

Figure: Determine unknown parameter $k$ by numeric procedure
Considering the book you mention is from 1960s, I somehow feel that the author's original intention was not to do numeric simulations to solve this problem. As far as I know there is no analytical method to solve roots of "quadratic plus exponential" equations
$$c y^2 + y + 1 - e^y = 0$$
where $c = \frac{x_f}{g t_f^2}$ and $y = -t_f k$.

If you are actually trying to calculate the velocity when acceleration drops to zero
$$v_t = \lim_{t \to \infty} v(t) = \frac{g}{k}$$
the situation is not much different
$$v_t = \frac{g x_f}{g t_f - v_f}$$
because $v_f$ is also unknown and cannot be solved without knowing $k$. Note that knowing only $m$ or only $b$ would not help much because its their ratio is what it matters. However, no numeric procedure would be needed if velocity $v_f$ were known in addition to $x_f$ and $t_f$.
A: Since you know x=108m and t=5s you can evaluate for b/m and than find v. So you had the solution almost.
