How to Derive Free Fall Model with Drag Resistance (Velocity-Time)?

Solution 1

This solution involves trigonometric substitution and integrals, imaginary numbers, and hyperbolic identities (derivation made by Rigel Melaan).

$\frac{1}{a{v}^2-g}dv=dt$

Integrate both sides of the equation:

$\int \frac{1}{a{v}^2-g}dv=\int dt$

Factor out $-g$ from the denominator and then the constants:

$\int -\frac{1}{g(1-\frac{a{v}^2}{g})}dv=\int dt$

$-\frac{1}{g}\int \frac{1}{1-\frac{a{v}^2}{g}}dv=\int dt$

Apply substitution of variables by defining variables $u$ and $du$. The $⟹$ indicates that we can rearrange the expression into another expression.

$u=iv\sqrt{\frac{a}{g}}⟹u^2=-\frac{a{v}^2}{g}$

$\frac{du}{dv}=i\sqrt{\frac{a}{g}}⟹dv=\frac{1}{i\sqrt{\frac{a}{g}}}du$

Substitute it back to the previous expression:

$-\frac{1}{g}\int \frac{1}{1-\frac{a{v}^2}{g}}dv=\int dt$

$-\frac{1}{g}\int \frac{1}{u^2+1}\times \frac{1}{i\sqrt{\frac{a}{g}}}du=\int dt$

Factor out constants and use the following identities. Note that $\wedge$ means "implies." It indicates an identity.

$-\frac{1}{g}\times \frac{1}{i\sqrt{\frac{a}{g}}}\int \frac{1}{u^2+1}du=\int dt\wedge \frac{1}{i}=-i$

$\frac{i}{ag}\int \frac{1}{u^2+1}du=\int dt\wedge \int \frac{1}{u^2+1}du=\arctan u$

$\frac{i\arctan u}{\sqrt{ag}}+c=t$

Substitute $u$ back and use the $\mathrm{arctanh}$ identity:

$\frac{i\arctan iv\sqrt{\frac{a}{g}}}{\sqrt{ag}}+c=t\wedge i\arctan (iu)=-\mathrm{arctanh}(u)$

$\frac{-\mathrm{arctanh}\left(v\sqrt{\frac{a}{g}}\right)}{\sqrt{ag}}+c=t$

Substitute $a$ back:

$-\sqrt{\frac{m}{kg}}\mathrm{arctanh}\left(v\sqrt{\frac{k}{mg}}\right)+c=t$

Rearrange the equation to get the general solution to the ODE:

$v(t)=\sqrt{\frac{mg}{k}}\tanh (c\sqrt{\frac{kg}{m}}-t\sqrt{\frac{kg}{m}})$

Solution 2

This solution involves trigonometric substitution, natural logarithms, and hyperbolic identities.

$\frac{1}{a{v}^2-g}dv=dt$

Integrate both sides of the equation:

$\int \frac{1}{a{v}^2-g}dv=\int dt$

Apply integration by substitution; Let $v$ and $dv$ be expressed in terms of $x$:

$v=\sqrt{\frac{g}{a}}x$

$dv=\sqrt{\frac{g}{a}}dx$

$\int \frac{1}{a\left(\frac{g{x}^2}{a}\right)-g}\times \sqrt{\frac{g}{a}}dx=\int dt$

Simplify. Eliminate $a$ in denominator and factor out $g$ and other constants:

$\int \frac{1}{g{x}^2-g}\times \sqrt{\frac{g}{a}}dx=\int dt$

$\int \frac{1}{g(x^2-1)}\times \sqrt{\frac{g}{a}}dx=\int dt$

$\frac{1}{\sqrt{ag}}\int \frac{1}{x^2-1}dx=\int dt$

Resolve the left side into partial fractions:

$-\frac{1}{2\sqrt{ag}}\int [\frac{1}{1+x}+\frac{1}{1-x}]dx=\int dt$

Integrate both sides of the equation:

$-\frac{1}{2\sqrt{ag}}[\ln (1+x)-\ln (1-x)]+c=t$

Use properties of logarithms to simplify and apply identity:

$-\frac{1}{\sqrt{ag}}[\frac{1}{2}\ln \left(\frac{1+x}{1-x}\right)]+c=t\wedge \frac{1}{2}\ln \left(\frac{1+u}{1-u}\right)=\mathrm{arctanh}u)$

$-\frac{\mathrm{arctanh}x}{\sqrt{ag}}+c=t$

Substitute $v$ back to equation:

$x=\sqrt{\frac{a}{g}}v$

$-\frac{1}{\sqrt{ag}}\mathrm{arctanh}\sqrt{\frac{a}{g}}v+c=t$

$a=\frac{k}{m}$

$-\sqrt{\frac{m}{kg}}\mathrm{arctanh}\left(v\sqrt{\frac{k}{mg}}\right)+c=t$

Rearrange the equation to get the general solution:

$v(t)=\sqrt{\frac{mg}{k}}\tanh (c\sqrt{\frac{kg}{m}}-t\sqrt{\frac{kg}{m}})$

Particular Solution

To find its particular solution, we let $v=v_o$ when $t=0$ to simulate free fall body:

$v(0)=\sqrt{\frac{mg}{k}}\tanh \left(c\sqrt{\frac{kg}{m}}\right)$

$c=\sqrt{\frac{m}{kg}}\mathrm{arctanh}\left(v_0\sqrt{\frac{k}{mg}}\right)$

Substitute $c$ back to the general solution:

$v(t)=\sqrt{\frac{mg}{k}}\tanh (\mathrm{arctanh}\left(v_0\sqrt{\frac{k}{mg}}\right)-t\sqrt{\frac{kg}{m}})$