A heavy point of mass m begins to fall in a medium with…
A heavy point of mass $m$ begins to fall in a medium with resistance $f=kv^n$. Determine the velocity as a function of time and the distance fallen in time $t$, for $n=1$ and $n=2$.
Explanation of Method
Apply Newton's second law to get $m\,dv/dt=mg-kv^n$, then solve the resulting separable ODE for each value of $n$ (a linear case and a Riccati-type case), integrating again to get the distance $s(t)=\int_0^tv\,d\tau$.
Step 1: Set up the equation of motion
With $\alpha=k/m$ and $v(0)=0$,
\[\frac{dv}{dt}=g-\alpha v^n.\]
Step 2: Case $n=1$ (linear resistance)
Separating $\dfrac{dv}{g-\alpha v}=dt$ and applying $v(0)=0$ gives
\[\boxed{\,v(t)=\frac{mg}{k}\Big(1-e^{-\frac{k}{m}t}\Big).}\]
Integrating $v$ from $0$ to $t$:
\[\boxed{\,s(t)=\frac{mg}{k}\,t-\frac{m^2g}{k^2}\Big(1-e^{-\frac{k}{m}t}\Big).}\]
Step 3: Case $n=2$ (quadratic resistance)
With terminal velocity $v_\infty=\sqrt{g/\alpha}$, the equation $\dfrac{dv}{dt}=g\Big(1-\dfrac{v^2}{v_\infty^2}\Big)$ separates via partial fractions; applying $v(0)=0$ gives
\[\boxed{\,v(t)=\sqrt{\frac{mg}{k}}\,\tanh\!\left(\sqrt{\frac{gk}{m}}\,t\right).}\]
Integrating,
\[\boxed{\,s(t)=\frac{m}{k}\ln\!\left[\cosh\!\left(\sqrt{\frac{gk}{m}}\,t\right)\right].}\]
Result
Final answer:
For $n=1$: $v(t)=\dfrac{mg}{k}\big(1-e^{-kt/m}\big)$, $\ s(t)=\dfrac{mg}{k}t-\dfrac{m^2g}{k^2}\big(1-e^{-kt/m}\big)$.
For $n=2$: $v(t)=\sqrt{\dfrac{mg}{k}}\tanh\!\Big(\sqrt{\dfrac{gk}{m}}\,t\Big)$, $\ s(t)=\dfrac{m}{k}\ln\!\Big[\cosh\!\Big(\sqrt{\dfrac{gk}{m}}\,t\Big)\Big]$.