Question:
A ball is thrown and it tracks a projectile near the Earth's surface subject to a resistive force due to air, $f_R = -bv$, which is proportional to its velocity. Find the velocities, positions for $x$ and $y$ directions; and the time of flight.
Answer:
Newton's second law shows
\begin{eqnarray}
mv'_x &=& -bv_x \\
mv'_y &=& -mg-bv_y
\end{eqnarray}
Solve the equation of motion in $x$ direction.
\begin{equation}
v_x = v_{0x}e^{-\frac{b}{m}t} \quad \rightarrow \quad v_x = v_{0x}e^{-\gamma t}
\end{equation}
where $\gamma=\frac{b}{m}$.
Integrate it again with respect to time.
\begin{eqnarray}
x &=& v_{0x}\int e^{-\gamma t} dt \\
&=& v_{0x} \left[ -\frac{1}{\gamma}e^{-\gamma t}\right]^t_0 \\
&=& \frac{v_{0x}}{\gamma}(1-e^{-\gamma t})
\end{eqnarray}
Likewise, we can integrate the $y$ component of the equation.
\begin{equation}
\frac{dv_y}{dt} = -g - \gamma v_y
\end{equation}
Then, we separate the terms for $v$ and the other.
\begin{equation}
\frac{dv_y}{dt} = - \gamma \left(v_y +\frac{g}{\gamma}\right)
\end{equation}
Now, divide it by $v_y +\frac{g}{\gamma}$ and multiply by $dt$ although this statement is not recommended for talking to mathematicians.
\begin{eqnarray}
& & \frac{dv}{v_y + \frac{g}{\gamma}} = -\gamma dt \\
& & \ln\left| v_y +\frac{g}{\gamma}\right| = -\gamma t + C \\
& & v_y + \frac{g}{\gamma} = e^{-\gamma t + C}\\
& & v_y = Ce^{-\gamma t} - \frac{g}{\gamma}
\end{eqnarray}
Consider the initial condition: When $t=0$, $v_y = v_{0y}$. Therefore, $C=v_{0y}+\frac{g}{\gamma}$. We now have
\begin{equation}
v_y = \left(v_{0y}+\frac{g}{\gamma}\right)e^{-\gamma t} - \frac{g}{\gamma}
\end{equation}
Integrate it again with time.
\begin{eqnarray}
y &=& \int \left\{\left(v_{0y}+\frac{g}{\gamma}\right)e^{-\gamma t} - \frac{g}{\gamma}\right\} dt \\
&=& \left[ -\frac{1}{\gamma}\left(v_{0y}+\frac{g}{\gamma}\right)e^{-\gamma t} - \frac{gt}{\gamma}\right]^t_0 \\
&=& \frac{1}{\gamma}\left(v_{0y}+\frac{g}{\gamma}\right)(1-e^{-\gamma t}) - \frac{gt}{\gamma}
\end{eqnarray}
When $y=0$, we can obtain the entire time of flight. Thus,
\begin{eqnarray}
\frac{gt}{\gamma} &=& \frac{1}{\gamma}\left(v_{0y}+\frac{g}{\gamma}\right)(1-e^{-\gamma t}) \\
\frac{g}{\gamma} &=& \left(v_{0y}+\frac{g}{\gamma}\right)\frac{1-e^{-\gamma t}}{\gamma t} \\
\frac{1-e^{-\gamma t}}{\gamma t} &=& \frac{g}{\gamma\left( v_{0y}+ \frac{g}{\gamma}\right)} \\
\frac{1-e^{-\gamma t}}{\gamma t} &=& \frac{1}{1+ \frac{\gamma v_{0y}}{g}}
\end{eqnarray}
Expand both sides by assuming that $\gamma \ll 1$.
\begin{equation}
{\textstyle 1-\frac{1}{2!}\gamma t+\frac{1}{3!}(\gamma t)^2-\frac{1}{4!}(\gamma t)^3+\cdots = 1-\frac{\gamma v_{0y}}{g}+\left(\frac{\gamma v_{0y}}{g}\right)^2-\left(\frac{\gamma v_{0y}}{g}\right)^3 + \cdots}
\end{equation}
Let $G=\frac{\gamma v_{0y}}{g}$, and expand $\gamma t$ in terms of polynomial of $G$. Namely,
\begin{equation}
\gamma t=a_1 G+a_2 G^2 + a_3 G^3 \cdots
\end{equation}
Plug this in the left hand side of (21). Let us take it up to $a_2$ and compare them with the right hand side in terms of $G^n$. Then, we have $a_1 = 2$ and $a_2 = -\frac{2}{3}$. Again, plug this in (22).
\begin{equation}
\gamma t = \frac{2v_{0y} \gamma}{g}-\frac{2}{3}\frac{\gamma^2 v_{0y}^2}{g^2}+\cdots
\end{equation}
Therefore the time $t=T$ is approximately given as follows:
\[
T = \frac{2v_{0y}}{g}-\frac{2}{3}\frac{\gamma v_{0y}^2}{g^2}
\]
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