### Solution 1

This solution involves trigonometric substitutions (derivation made by Rigel Melaan).

Express instantaneous velocity as the derivative of position with respect to time:

\(v(t)=\lim _{\Delta t \rightarrow 0} \frac{\Delta s}{\Delta t}=\frac{d s}{d t}\)

Rearrange the equation and introduce the v-t model:

\(s(t)=\int v d t=\)

\(\int \sqrt{\frac{m g}{k}} \tanh \left(\operatorname{arctanh}\left(v_0 \sqrt{\frac{k}{m g}}\right)-t \sqrt{\frac{k g}{m}}\right) d t\)

Factor out the constants:

\(s(t)=\sqrt{\frac{m g}{k}} \int \tanh \left(\operatorname{arctanh}\left(v_0 \sqrt{\frac{k}{m g}}\right)-t \sqrt{\frac{k g}{m}}\right) d t\)

Introduce variables \(u\) and \(du\):

\(u=\operatorname{arctanh}\left(v_0 \sqrt{\frac{k}{m g}}\right)-t \sqrt{\frac{k g}{m}}\)

\(\frac{d u}{d t}=-\sqrt{\frac{k g}{m}} \Longrightarrow d t=-\sqrt{\frac{m}{k g}} d u\)

Substitute \(u\) and \(du\) and simplify:

\(s(t)=\sqrt{\frac{m g}{k}} \int \tanh (u) \times-\sqrt{\frac{m}{k g}} d u\)

\(s(t)=-\frac{m}{k} \int \tanh (u) d u\)

Define \(v\) and \(dv\):

\(v=\cosh (u)\)

\(\frac{d v}{d u}=\sinh u \Longrightarrow d u=\frac{1}{\sinh u} \times d v\)

Use identity of \(\sinh\) and substitute:

\(s(t)=-\frac{m}{k} \int \frac{\sinh u}{v} \frac{1}{\sinh u} d v=\)

\(-\frac{m}{k} \int \frac{1}{v} d v=-\frac{m}{k} \ln |v|+c\)

Substitute \(u\) back to the model. At time \(t=0\), \(c\) is \(s_0\):

\(s(t)=s_0-\frac{m \ln \left(\cosh \left(\operatorname{arctanh}\left(v_0 \sqrt{\frac{k}{m g}}\right)-t \sqrt{\frac{k g}{m}}\right)\right)}{k}\)