A body falls from rest under gravity through a resisting…
A body falls from rest under gravity through a resisting medium whose resistance is proportional to the square of the velocity. Given that the terminal velocity is $v_\infty = 75\ \text{m/s}$ and $g=9.81\ \text{m/s}^2$, find the velocity $v(t)$ and the distance fallen $x(t)$ as functions of time.
Explanation of Method
Resistance proportional to v², separable in the velocity, giving hyperbolic-tangent motion and a logarithmic-cosh displacement.
Step 1: Setting up the equation of motion
Let $v=\dot x$ be the downward velocity. With a resisting force proportional to $v^2$, Newton's second law reads
\[ m\dot v = mg - kv^2. \]
At the terminal velocity $v_\infty$, the acceleration vanishes, so $mg = kv_\infty^2$, i.e. $k = mg/v_\infty^2$. Substituting this back, $\dot v = g\left(1 - \frac{v^2}{v_\infty^2}\right)$.
Step 2: Solving for the velocity
Separating variables, $\frac{v_\infty^2\,dv}{v_\infty^2 - v^2} = g\,dt$.
Using the partial-fraction decomposition
\[ \frac{v_\infty^2}{v_\infty^2-v^2} = \frac{v_\infty}{2}\left(\frac{1}{v_\infty-v} + \frac{1}{v_\infty+v}\right), \]
integrating both sides and applying $v(0)=0$ (which makes the constant of integration vanish) gives
\[ \frac{v_\infty}{2}\ln\!\left(\frac{v_\infty+v}{v_\infty-v}\right) = gt. \]
Writing $\alpha = g/v_\infty$ and solving for $v$, \[\frac{v_\infty+v}{v_\infty-v} = e^{2\alpha t} \quad\Longrightarrow\quad v(t) = v_\infty\tanh(\alpha t) = v_\infty\tanh\!\left(\frac{gt}{v_\infty}\right).\]
Step 3: Solving for the position
Integrating $v(t)$ with $x(0)=0$, \[x(t) = \int_0^t v_\infty\tanh\!\left(\frac{g\tau}{v_\infty}\right)d\tau = \frac{v_\infty^2}{g}\ln\cosh\!\left(\frac{gt}{v_\infty}\right).\]
Step 4: The laws of motion
\[v(t) = v_\infty\tanh\!\left(\frac{gt}{v_\infty}\right), \qquad x(t) = \frac{v_\infty^2}{g}\ln\cosh\!\left(\frac{gt}{v_\infty}\right)\]
Direct substitution confirms both: $\dot v - g(1-v^2/v_\infty^2)$ reduces identically to zero, and $\dot x - v$ reduces identically to zero.
Step 5: Numerical evaluation for $v_\infty=75\ \text{m/s}$, $g=9.81\ \text{m/s}^2$
Here $g/v_\infty \approx 0.1308\ \text{s}^{-1}$ and $v_\infty^2/g \approx 573.4\ \text{m}$, so
\[ v(t) \approx 75\tanh(0.1308\,t)\ \text{m/s}, \qquad x(t) \approx 573.4\,\ln\cosh(0.1308\,t)\ \text{m}. \]
Result
Final answer:
\[\boxed{\,v(t) = v_\infty\tanh\!\left(\frac{gt}{v_\infty}\right), \qquad x(t) = \frac{v_\infty^2}{g}\ln\cosh\!\left(\frac{gt}{v_\infty}\right)\,}\]
Remark
Remark. As $t\to\infty$, $\tanh(gt/v_\infty)\to 1$, so $v(t)\to v_\infty=75\ \text{m/s}$, as required. For short times, when $gt/v_\infty\ll1$, the approximations $\tanh u\approx u$ and $\ln\cosh u\approx u^2/2$ give $v(t)\approx gt$ and $x(t)\approx \tfrac12 gt^2$ — recovering ordinary unresisted free fall, as expected when resistance is still negligible. For large times, $x(t)\approx v_\infty t - \dfrac{v_\infty^2\ln 2}{g}$: the motion becomes asymptotically linear at the terminal speed, offset by a constant.