What are the formulae for resisted projectile motion in which the resistance is proportional to the velocity? I have a problem where my answers don't match up with the textbook answers and I need to verify with formulae.
I need the velocities and positions with respect to time for both components, for future readers as well. The projectile is launched from the ground, at an angle $\theta$ and velocity $u$:
Best Answer
We can start with the initial velocities:
$$ \begin{align} \dot y(0) &= u\sin\theta \\ \dot x(0) &= u\cos\theta \end{align} $$
The drag coefficient is $k$ so we can write the forces as:
$$ \begin{align} m\ddot y &= -mg -k\dot y \\ m\ddot x &= -k\dot x \end{align} $$
Resolving the equations w.r.t. the acceleration yields:
$$ \begin{align} \ddot y &= -g - \frac{k}{m} \dot y \\ \ddot x &= - \frac{k}{m} \dot x \end{align} $$
Rewrite the acceleration in terms of time:
$$ \begin{align} \frac{dt}{d \dot y} &= - \frac{1}{g + \frac{k}{m} \dot y} \\ \frac{dt}{d \dot x} &= -\frac{m}{k} \cdot \frac{1}{\dot x} \end{align} $$
Integrating:
$$ \begin{align} t &= - \frac{m}{k}\ln(g + \frac{k}{m}\dot y) + C \\ t &= - \frac{m}{k}\ln(\dot x) + C \end{align} $$
Substituting the initial velocities $\dot x(0)$ and $\dot y(0)$ at $t=0$: $$ \begin{align} t &= \frac{m}{k}\ln(\frac{g + \frac{k}{m}\dot y(0)}{g+\frac{k}{m}\dot y}) \\ t &= \frac{m}{k}\ln(\frac{\dot x(0)}{\dot x}) \end{align} $$
Rewrite in terms of $\dot x$ or $\dot y$
$$ \begin{align} \dot y &= [\frac{m}{k} g + \dot y(0)]e^{-\frac{k}{m}t} - \frac{m}{k}g \\ \dot x &= \dot x(0) e^{-\frac{k}{m}t} \\ \end{align} $$
We can integrate again to find the positions.