Formulae for projectile motion with resistance proportional to velocity 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 $u$:
 A: 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.
