So, this is basically the same situation as in the previous example. We just changed the air resistance from \(5v\) to \(5{v^2}\). Also, we are just going to find the velocity at any time \(t\) for this problem because, we’ll the solution is really unpleasant and finding the velocity for when the mass hits the ground is simply more work that we want to put into a problem designed to illustrate the fact that we need a separate differential equation for both the upwards and downwards motion of the mass.

As with the previous example we will use the convention that everything downwards is positive. Here are the forces on the mass when the object is on the way and on the way down.

Now, in this case, when the object is moving upwards the velocity is negative. However, because of the \({v^2}\) in the air resistance we do not need to add in a minus sign this time to make sure the air resistance is positive as it should be given that it is a downwards acting force.

On the downwards phase, however, we still need the minus sign on the air resistance given that it is an upwards force and so should be negative but the \({v^2}\) is positive.

This leads to the following IVP’s for each case.

\[\begin{array}{*{20}{c}}\begin{aligned}&\hspace{0.5in}{\mbox{Up}}\\ & mv' = mg + 5{v^2}\\ & v' = 9.8 + \frac{1}{{10}}{v^2}\\ & v\left( 0 \right) = - 10\end{aligned}&\begin{aligned}&\hspace{0.35in}{\mbox{Down}}\\ & mv' = mg - 5{v^2}\\ & v' = 9.8 - \frac{1}{{10}}{v^2}\\ & v\left( {{t_0}} \right) = 0\end{aligned}\end{array}\]

These are clearly different differential equations and so, unlike the previous example, we can’t just use the first for the full problem.

Also, the solution process for these will be a little more involved than the previous example as neither of the differential equations are linear. They are both separable differential equations however.

So, let’s get the solution process started. We will first solve the upwards motion differential equation. Here is the work for solving this differential equation.

\[\begin{align*}\int{{\frac{1}{{9.8 + \frac{1}{{10}}{v^2}}}\,dv}} & = \int{{dt}}\\ 10\int{{\frac{1}{{98 + {v^2}}}\,dv}} & = \int{{dt}}\\ \frac{{10}}{{\sqrt {98} }}{\tan ^{ - 1}}\left( {\frac{v}{{\sqrt {98} }}} \right) & = t + c\end{align*}\]

Note that we did a little rewrite on the integrand to make the process a little easier in the second step.

Now, we have two choices on proceeding from here. Either we can solve for the velocity now, which we will need to do eventually, or we can apply the initial condition at this stage. While, we’ve always solved for the function before applying the initial condition we could just as easily apply it here if we wanted to and, in this case, will probably be a little easier.

So, to apply the initial condition all we need to do is recall that \(v\) is really \(v\left( t \right)\) and then plug in \(t = 0\). Doing this gives,

\[\frac{{10}}{{\sqrt {98} }}{\tan ^{ - 1}}\left( {\frac{{v\left( 0 \right)}}{{\sqrt {98} }}} \right) = 0 + c\]

Now, all we need to do is plug in the fact that we know \(v\left( 0 \right) = - 10\) to get.

\[c = \frac{{10}}{{\sqrt {98} }}{\tan ^{ - 1}}\left( {\frac{{ - 10}}{{\sqrt {98} }}} \right)\]

Messy, but there it is. The velocity for the upward motion of the mass is then,

\[\begin{align*}\frac{{10}}{{\sqrt {98} }}{\tan ^{ - 1}}\left( {\frac{v}{{\sqrt {98} }}} \right) & = t + \frac{{10}}{{\sqrt {98} }}{\tan ^{ - 1}}\left( {\frac{{ - 10}}{{\sqrt {98} }}} \right)\\ {\tan ^{ - 1}}\left( {\frac{v}{{\sqrt {98} }}} \right) & = \frac{{\sqrt {98} }}{{10}}t + {\tan ^{ - 1}}\left( {\frac{{ - 10}}{{\sqrt {98} }}} \right)\\ v\left( t \right) & = \sqrt {98} \tan \left( {\frac{{\sqrt {98} }}{{10}}t + {{\tan }^{ - 1}}\left( {\frac{{ - 10}}{{\sqrt {98} }}} \right)} \right)\end{align*}\]

Now, we need to determine when the object will reach the apex of its trajectory. To do this all we need to do is set this equal to zero given that the object at the apex will have zero velocity right before it starts the downward motion. We will leave it to you to verify that the velocity is zero at the following values of \(t\).

\[t = \frac{{10}}{{\sqrt {98} }}\left[ {{{\tan }^{ - 1}}\left( {\frac{{10}}{{\sqrt {98} }}} \right) + \pi n} \right]\hspace{0.25in}n = 0, \pm 1, \pm 2, \pm 3, \ldots \]

We clearly do not want all of these. We want the first positive \(t\) that will give zero velocity. It doesn’t make sense to take negative \(t\)’s given that we are starting the process at \(t = 0\) and once it hit’s the apex (*i.e.* the first positive \(t\) for which the velocity is zero) the solution is no longer valid as the object will start to move downwards and this solution is only for upwards motion.

Plugging in a few values of \(n\) will quickly show us that the first positive \(t\) will occur for \(n = 0\) and will be \(t = 0.79847\). We reduced the answer down to a decimal to make the rest of the problem a little easier to deal with.

The IVP for the downward motion of the object is then,

\[v' = 9.8 - \frac{1}{{10}}{v^2}\hspace{0.25in}v\left( {0.79847} \right) = 0\]

Now, this is also a separable differential equation, but it is a little more complicated to solve. First, let’s separate the differential equation (with a little rewrite) and at least put integrals on it.

\[\int{{\frac{1}{{9.8 - \frac{1}{{10}}{v^2}}}\,dv}} = 10\int{{\frac{1}{{98 - {v^2}}}\,dv}} = \int{{dt}}\]

The problem here is the minus sign in the denominator. Because of that this is not an inverse tangent as was the first integral. To evaluate this integral we could either do a trig substitution (\(v = \sqrt {98} \sin \theta \)) or use partial fractions using the fact that \(98 - {v^2} = \left( {\sqrt {98} - v} \right)\left( {\sqrt {98} + v} \right)\). You’re probably not used to factoring things like this but the partial fraction work allows us to avoid the trig substitution and it works exactly like it does when everything is an integer and so we’ll do that for this integral.

We’ll leave the details of the partial fractioning to you. Once the partial fractioning has been done the integral becomes,

\[\begin{align*}10\left( {\frac{1}{{2\sqrt {98} }}} \right)\int{{\frac{1}{{\sqrt {98} + v}} + \frac{1}{{\sqrt {98} - v}}\,dv}} & = \int{{dt}}\\ \frac{5}{{\sqrt {98} }}\left[ {\ln \left| {\sqrt {98} + v} \right| - \ln \left| {\sqrt {98} - v} \right|} \right] & = t + c\\ \frac{5}{{\sqrt {98} }}\ln \left| {\frac{{\sqrt {98} + v}}{{\sqrt {98} - v}}} \right| & = t + c\end{align*}\]

Again, we will apply the initial condition at this stage to make our life a little easier. Doing this gives,

\[\begin{align*}\frac{5}{{\sqrt {98} }}\ln \left| {\frac{{\sqrt {98} + v\left( {0.79847} \right)}}{{\sqrt {98} - v(0.79847}}} \right| & = 0.79847 + c\\ \frac{5}{{\sqrt {98} }}\ln \left| {\frac{{\sqrt {98} + 0}}{{\sqrt {98} - 0}}} \right| & = 0.79847 + c\\ \frac{5}{{\sqrt {98} }}\ln \left| 1 \right| & = 0.79847 + c\\ c & = - 0.79847\end{align*}\]

The solutions, as we have it written anyway, is then,

\[\frac{5}{{\sqrt {98} }}\ln \left| {\frac{{\sqrt {98} + v}}{{\sqrt {98} - v}}} \right| = t - 0.79847\]

Okay, we now need to solve for \(v\) and to do that we really need the absolute value bars gone and no we can’t just drop them to make our life easier. We need to know that they can be dropped without have any effect on the eventual solution.

To do this let’s do a quick direction field, or more appropriately some sketches of solutions from a direction field. Here is that sketch,

Note that \(\sqrt {98} = 9.89949\) and so is slightly above/below the lines for -10 and 10 shown in the sketch. The important thing here is to notice the middle region. If the velocity starts out anywhere in this region, as ours does given that \(v\left( {0.79847} \right) = 0\), then the velocity must always be less that \(\sqrt {98} \).

What this means for us is that both \(\sqrt {98} + v\) and \(\sqrt {98} - v\) must be positive and so the quantity in the absolute value bars must also be positive. Therefore, in this case, we can drop the absolute value bars to get,

\[\] \[\frac{5}{{\sqrt {98} }}\ln \left[ {\frac{{\sqrt {98} + v}}{{\sqrt {98} - v}}} \right] = t - 0.79847\]

At this point we have some very messy algebra to solve for \(v\). We will leave it to you to verify our algebra work. The solution to the downward motion of the object is,

\[v\left( t \right) = \sqrt {98} \frac{{{{\bf{e}}^{\frac{1}{5}\sqrt {98} \left( {t - 0.79847} \right)}} - 1}}{{{{\bf{e}}^{\frac{1}{5}\sqrt {98} \left( {t - 0.79847} \right)}} + 1}}\]

Putting everything together here is the full (decidedly unpleasant) solution to this problem.

\[v\left( t \right) = \left\{ {\begin{array}{ll}{\sqrt {98} \tan \left( {\frac{{\sqrt {98} }}{{10}}t + {{\tan }^{ - 1}}\left( {\frac{{ - 10}}{{\sqrt {98} }}} \right)} \right)}&{0 \le t \le 0.79847\,\,\,\left( {{\mbox{upward motion}}} \right)}\\{\sqrt {98} \frac{{{{\bf{e}}^{\frac{1}{5}\sqrt {98} \left( {t - 0.79847} \right)}} - 1}}{{{{\bf{e}}^{\frac{1}{5}\sqrt {98} \left( {t - 0.79847} \right)}} + 1}}}&{0.79847 \le t \le {t_{{\mathop{\rm end}\nolimits} }}\,\,\left( {{\mbox{downward motion}}} \right)}\end{array}} \right.\]

where \({t_{{\mbox{end}}}}\) is the time when the object hits the ground. Given the nature of the solution here we will leave it to you to determine that time if you wish to but be forewarned the work is liable to be very unpleasant.