Motion of Falling Body w/ Varying Drag: Farah's Q on Yahoo Answers

[MarkFL]

Here is the question:

Luke Skywalker (m=100kg) fell from rest off the antenna below the cloud city of Bespin (height=100000 m).

Assume that he missed the Millennium Falcon waiting below and continued to plummet. Suppose the gravitational acceleration of Bespin is 10 m/s^2, and the force of drag provides resistance proportional to the magnitude of his velocity, with a drag coefficient of k=4. This question is also from my differential class thank you in advance.

Additional Details:

a) Find Luke's velocity and displacement.

b) Find Luke's limiting velocity.

c) Suppose it would take 200 seconds for the millennium falcon to shake of the pursuing TIE fighters and turn around for a second rescue attempt. How far from the ground would the rescue be made?

I have posted a link there to this topic so the OP can see my work.

 

[MarkFL]

Re: Farah's question at Yahoo! Answers ragarding motion of a falling body where drag varies as veloc

Hello Farah,

I would begin with Newton's second law of motion:

\(\displaystyle ma=F\)

Since acceleration $a$ is defined to be the time rate of change of velocity $v$, we may write:

\(\displaystyle m\frac{dv}{dt}=F\)

We have two forces acting on Luke...the force of gravity and drag, which we are told is proportional to his velocity, hence:

\(\displaystyle m\frac{dv}{dt}=mg-kv\)

We subtract the force due to drag since it opposes the motion. Treating $dv$ and $dt$ as differentials, we may separate variables to obtain:

\(\displaystyle \frac{dv}{mg-kv}=\frac{dt}{m}\)

Integrating we find:

\(\displaystyle \int\frac{1}{mg-kv}\,dv=\frac{1}{m}\int\,dt\)

\(\displaystyle \ln(mg-kv)=C-\frac{kt}{m}\)

Converting from logarithmic to exponential form, we find:

\(\displaystyle mg-kv=Ce^{-\frac{kt}{m}}\) where \(\displaystyle 0<C\)

Solving for $v$, we obtain:

\(\displaystyle v(t)=\frac{mg}{k}-Ce^{-\frac{kt}{m}}\)

Now, we are told Luke begins at rest, which means \(\displaystyle v(0)=0\), and so from this initial value, we may determine the parameter $C$:

\(\displaystyle v(0)=\frac{mg}{k}-C=0\,\therefore\,C=\frac{mg}{k}\)

And so the solution satisfying the IVP is:

\(\displaystyle v(t)=\frac{mg}{k}-\frac{mg}{k}e^{-\frac{kt}{m}}=\frac{mg}{k}\left(1-e^{-\frac{kt}{m}} \right)\)

With $k=4\,\dfrac{\text{kg}}{\text{s}},\,m=100\text{ kg},\,g=10\,\dfrac{\text{m}}{\text{s}^2}$, we then have:

\(\displaystyle v(t)=250\left(1-e^{-\frac{t}{25}} \right)\)

Next, we may determine Luke's position $x(t)$ above the ground by using the definition:

\(\displaystyle \frac{dx}{dt}\equiv v(t)\)

If we orient our vertical $x$-axis such that the origin is on the ground and the positive direction is up, or opposing velocity, we then have:

\(\displaystyle \frac{dx}{dt}=250\left(e^{-\frac{t}{25}}-1 \right)\)

Separating variables, we obtain:

\(\displaystyle dx=250\left(e^{-\frac{t}{25}}-1 \right)\,dt\)

and integrating:

\(\displaystyle \int\,dx=250\int e^{-\frac{t}{25}}-1\,dt\)

\(\displaystyle x(t)=250\left(-25e^{-\frac{t}{25}}-t \right)+C\)

Now using the fact that $x(0)=100000$, we may use this to determine the parameter $C$:

\(\displaystyle x(0)=250\left(-25 \right)+C=100000\,\therefore C=106250\)

And so the solution satisfying the IVP is:

\(\displaystyle x(t)=250\left(-25e^{-\frac{t}{25}}-t \right)+106250=250\left(-25e^{-\frac{t}{25}}-t+425 \right)\)

Now we may answer the questions:

a) Find Luke's velocity and displacement.

\(\displaystyle v(t)=250\left(1-e^{-\frac{t}{25}} \right)\)

\(\displaystyle x(t)=250\left(-25e^{-\frac{t}{25}}-t+425 \right)\)

b) Find Luke's limiting velocity.

\(\displaystyle \lim_{t\to\infty}v(t)=250\,\frac{\text{m}}{\text{s}}\)

c) Suppose it would take 200 seconds for the millennium falcon to shake of the pursuing TIE fighters and turn around for a second rescue attempt. How far from the ground would the rescue be made?

\(\displaystyle x(200)\approx56247.9033585756\text{ m}\)