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 with a constant \(b\). Find the velocities and positions for \(x\) and \(y\) directions; and the time of flight.

Answer:

Newton’s second law shows in each direction:

\[  mv’_x = -bv_x  \]

\[  mv’_y = -mg-bv_y  \]

Solve the equation of motion in \(x\) direction.

\[  v_x = v_{0x}e^{-\frac{b}{m}t} \quad \rightarrow \quad v_x = v_{0x}e^{-\gamma t}  \]

where \(\gamma=\frac{b}{m}\).

Integrate it again with respect to time.

\[  x = v_{0x}\int e^{-\gamma t} dt  \]

\[   = v_{0x} \left[ -\frac{1}{\gamma}e^{-\gamma t}\right]^t_0  \]

\[  = \frac{v_{0x}}{\gamma}(1-e^{-\gamma t})  \]

Likewise, we can integrate the \(y\) component of the equation.

\[  \frac{dv_y}{dt} = -g  – \gamma v_y  \]

Then, we separate the terms for \(v\) and the other as follows:

\[  \frac{dv_y}{dt} = – \gamma \left(v_y +\frac{g}{\gamma}\right)  \]

Now, divide it by \(v_y +\frac{g}{\gamma}\) and multiply by \(dt\) technically.

\[   \frac{dv}{v_y + \frac{g}{\gamma}} = -\gamma dt  \]

\[   \ln\left| v_y +\frac{g}{\gamma}\right| = -\gamma t + C \]

\[   v_y + \frac{g}{\gamma} = e^{-\gamma t + C}  \]

\[   v_y = Ce^{-\gamma t} – \frac{g}{\gamma}  \]

Consider the initial condition: When \(t=0\), \(v_y = v_{0y}\). Therefore, \(C=v_{0y}+\frac{g}{\gamma}\). We now have

\[  v_y = \left(v_{0y}+\frac{g}{\gamma}\right)e^{-\gamma t} – \frac{g}{\gamma}  \]

Integrate it again with time.

\[  y = \int \left\{\left(v_{0y}+\frac{g}{\gamma}\right)e^{-\gamma t} – \frac{g}{\gamma}\right\} dt \]

\[  = \left[ -\frac{1}{\gamma}\left(v_{0y}+\frac{g}{\gamma}\right)e^{-\gamma t} – \frac{gt}{\gamma}\right]^t_0  \]

\[  = \frac{1}{\gamma}\left(v_{0y}+\frac{g}{\gamma}\right)(1-e^{-\gamma t}) – \frac{gt}{\gamma}  \]

When \(y=0\), we can obtain the entire time of flight. Thus,

\[  \frac{gt}{\gamma} = \frac{1}{\gamma}\left(v_{0y}+\frac{g}{\gamma}\right)(1-e^{-\gamma t}) \]

\[  \frac{g}{\gamma} = \left(v_{0y}+\frac{g}{\gamma}\right)\frac{1-e^{-\gamma t}}{\gamma t} \]

\[  \frac{1-e^{-\gamma t}}{\gamma t} = \frac{g}{\gamma\left( v_{0y}+ \frac{g}{\gamma}\right)} \]

\[  \frac{1-e^{-\gamma t}}{\gamma t} = \frac{1}{1+ \frac{\gamma v_{0y}}{g}}  \]

Expand both sides by assuming that \(\gamma \ll 1\).

\[  {\textstyle 1-\frac{1}{2!}\gamma t+\frac{1}{3!}(\gamma t)^2-\frac{1}{4!}(\gamma t)^3+\cdots = 1-\frac{\gamma v_{0y}}{g}+\left(\frac{\gamma v_{0y}}{g}\right)^2-\left(\frac{\gamma v_{0y}}{g}\right)^3 + \cdots}  … (a)  \]

Let \(G=\frac{\gamma v_{0y}}{g}\), and expand \(\gamma t\) in terms of polynomial of \(G\). Namely,

\[  \gamma t=a_1 G+a_2 G^2 + a_3 G^3 \cdots  …(b) \]

Plug this in the left hand side of (a). 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 (b).

\[  \gamma t = \frac{2v_{0y} \gamma}{g}-\frac{2}{3}\frac{\gamma^2 v_{0y}^2}{g^2}+\cdots  \]

Therefore the time \(t=T\) is approximately given as follows:

\[  T = \frac{2v_{0y}}{g}-\frac{2}{3}\frac{\gamma v_{0y}^2}{g^2}  \]