How can I solve $\beta^2=\frac{m^2g}{h}\left(-\frac{\beta t}{m}+e^{\frac{\beta t}{m}}-1\right)$ for $\beta$?

Question

This equation arose when I tried to find out how to derive $\beta$ in Stokes' Drag Force $F=\beta v$ as a function of the time $t$ it takes a mass $m$ to hit the ground after falling from a height $h$: $$\beta^2=\dfrac{m^2g}{h}\left(-\dfrac{\beta t}{m}+e^{\frac{\beta t}{m}}-1\right)$$ I can't figure out how to solve it.

Answer 1

Let's take the physics approach, accepting that our model isn't perfectly descriptive of reality, and therefore solving this equation exactly isn't really worth the effort. Then we simply choose to drop the mass from a height $h$ such that the time it hits the ground $t$ satisfies $\beta t/m \ll 1$. Then Taylor expand the exponential to obtain \begin{align} \beta^{2} = \frac{m^{2}g}{h}\left((\beta t/m)^{2}/2 + (\beta t/m)^{3}/6 \right) + \mathcal{O}((\beta t/m)^4) \end{align} I'll let you take it from there.