# Appendix B:     Derivation of Terminal Velocity (Beyond Syllabus)

To obtain the exact formula for the velocity of a falling object, we must first model exactly how the drag force varies with velocity.

Laminar Flow

If laminar flow is assumed, then the drag force is proportional to speed ${{F}_{D}}=kv$

The terminal velocity is then given by $mg=k{{v}_{T}}\text{ }\Rightarrow \text{ }{{v}_{T}}=\frac{{mg}}{k}$.

Applying N2L to the falling object, we get \displaystyle \begin{aligned}{{F}_{{net}}}&=ma\\mg-kv&=m\frac{{dv}}{{dt}}\\1-\frac{k}{{mg}}v&=\frac{1}{g}\frac{{dv}}{{dt}}\\gdt&=\frac{1}{{1-\frac{k}{{mg}}v}}dv\end{aligned}

Integrating both sides, we get \displaystyle \begin{aligned}\int{g}\text{ }dt&=\int{{\frac{1}{{1-\frac{k}{{mg}}v}}\,dv}}\\gt&=-\frac{{mg}}{k}\ln (1-\frac{k}{{mg}}v)\\-\frac{k}{m}t&=\ln (1-\frac{k}{{mg}}v)\\{{e}^{{-\frac{k}{m}t}}}&=1-\frac{k}{{mg}}v\\v&=\frac{{mg}}{k}(1-{{e}^{{-\frac{k}{m}t}}})\\\text{ }v&={{v}_{T}}(1-{{e}^{{-\frac{k}{m}t}}})\end{aligned}

Turbulent Flow

If turbulent flow is assumed, then the drag force is proportional to the square of speed ${{F}_{D}}=k{{v}^{2}}$

The terminal velocity is then given by $mg=k{{v}_{T}}^{2}\text{ }\Rightarrow \text{ }{{v}_{T}}=\sqrt{{\frac{{mg}}{k}}}$.

Applying N2L to the falling object, we get \begin{aligned}{{F}_{{net}}}&=ma\\mg-k{{v}^{2}}&=m\frac{{dv}}{{dt}}\\\int{{\frac{1}{{1-\frac{{{{v}^{2}}}}{{{{v}_{T}}^{2}}}}}\,}}dv&=\int{g}\,dt\text{ }\Rightarrow \text{ }v={{v}_{T}}\tanh (\sqrt{{\frac{{kg}}{m}}}t)\end{aligned}

As you can see, the derivation involves quite a bit of integration (which is why it is not in the H2 syllabus), but the mathematics show that the terminal velocity is approached asymptotically, according to an exponential function under laminar flow conditions (e.g. tiny ball falling in honey), but hyperbolic tangent function under turbulent flow conditions (e.g. heavy ball falling in air).