velocity of falling object from different heights using DE

sunnyceej

1. The problem statement, all variables and given/known data
Find the velocity of a 180 lb object falling from a state of rest from 5 ft, 15 ft, 25 ft, and 45 ft

2. Relevant equations

using the formula f=mg-Kv², where g is the acceleration due to gravity, m the mass of the body, v its velocity. The terminal velocity is 120mph.

3. The attempt at a solution
I tried using f=m(dv/dt) so m(dv/dt)=mg-Kv², since mg = 180 and I know terminal velocity, I converted 120mph to fps and got k=1.614E-6. Using g=32fps², I got m=5.6. Then dividing by m, I get dv/dx=32 -1.614E-6v^2/5.6.

I've used Kinematic equations to get times at all of the heights, and from there I can get velocity, but none of this considers the drag, or uses the equation I was given.

I have no idea how to use the equation I was given to get the velocity for the different heights or times. The equation doesn't seem separable, or like I could use an integrating factor. I've tired Bernoulli's method too. None of them seem to help solve the problem.
Please help!

SammyS

Of course it's separable.

m(dv/dt)=mg-Kv² → [itex]\displaystyle \frac{dv}{mg-Kv^2}=\frac{dt}{m}\,.[/itex]

sunnyceej

If mg, and k =1, then I know I would get arctanh v. It's been a while since calc, so I'm not sure what to do with the constants.

My other thought is to go the natural log route:
ln(mg-kv²)=t/m+c
then raising both sides to e:
mg-kv²=ce^t/m
using v(0)=0 as my initial condition, c=180, and plugging in other values (mentioned in first post) gives me
180-1.614E^-6v²=180e^t/5.6

Am I on the right track?

If so, this gives me time, do I plug in my times from using the Kinematic equations to find the different velocities? My only hesitation here is that the times I found don't take into account the drag force. So, how do I account for the different heights? Or are the height and time interchangeable (can I replace t with h)?

sunnyceej

After thinking more, differentiating with respect to height would not change my equation (except from t to h), but my initial value would be different for each height. So I would have a different specific solution for each height.

I'm still curious as to whether I solved the equation correctly though.

SammyS

The constants can be taken care of through substitution.

While the integration can result in an arctanh function, it's fairly straight forward to express it as a sum of logarithms.

To get the integrand into workable form:

[itex]\displaystyle \frac{1}{mg-Kv^2}=\frac{1}{\displaystyle mg\left(1-\frac{K}{mg}v^2\right)}=\frac{1}{\displaystyle mg\left(1-\left(\sqrt{\frac{K}{mg}}v\right)^2\right)}[/itex]

[itex]\displaystyle \text{Let }u=\sqrt{\frac{K}{mg}}\,v\,,\text{ then }du=\sqrt{\frac{K}{mg}}\ dv\,.[/itex]

The integral then becomes:

[itex]\displaystyle \int \frac{dv}{mg-Kv^2}= \sqrt{\frac{1}{Kmg}}\ \int\frac{du}{1-u^2}\,.[/itex]

Notice that [itex]\displaystyle\frac{1}{1-u^2}=\frac{1}{2(1+u)}+\frac{1}{2(1-u)}\ .[/itex]

sunnyceej

So this gives me
[itex]\int\frac{dv}{mg-kv^{2}}[/itex] = [itex]\int\frac{dt}{m}[/itex]
with u substitution:
[itex]\sqrt{\frac{k}{mg}v}[/itex]([itex]\frac{1}{2}[/itex][itex]\int\frac{du}{1+u}[/itex] + [itex]\frac{1}{2}[/itex][itex]\int\frac{du}{1-u}[/itex]) = [itex]\int\frac{dt}{m}[/itex]

becomes

[itex]\sqrt{\frac{k}{mg}v}(\frac{1}{2}[/itex]ln[itex]\left|1+u\right|[/itex] + [itex]\frac{1}{2}[/itex]ln[itex]\left|1-u\right|[/itex]) = [itex]\frac{t}{m}[/itex]+c

then
[itex]\sqrt{\frac{k}{mg}v}(ln\sqrt{1+u}[/itex]+ln[itex]\sqrt{1-u}[/itex])= [itex]\frac{t}{m}[/itex]+c

[itex]\sqrt{\frac{k}{mg}v}(ln\sqrt{1-u^{2}}[/itex])=[itex]\frac{t}{m}[/itex]+c

and finally sub back in for u to get:
[itex]\sqrt{\frac{k}{mg}v}*ln\sqrt{1-\frac{k}{mg}v^{2}}[/itex]=[itex]\frac{t}{m}[/itex]+c

Is this right?

SammyS

That should be a minus sign.

[itex]\displaystyle\sqrt{\frac{k}{mg}v}(\frac{1}{2}[/itex]ln[itex]\left|1+u\right|[/itex] − [itex]\frac{1}{2}[/itex]ln[itex]\left|1-u\right|[/itex]) = [itex]\frac{t}{m}[/itex]+c

sunnyceej

Ok. I see why. I also now see why I should have 1/[itex]\sqrt{kmg}[/itex] in front instead of what I had.

So I now have [itex]\frac{1}{\sqrt{kmg}}[/itex][itex]\left(ln\sqrt{1+u}-ln\sqrt{1-u}\right)[/itex]=[itex]\frac{t}{m}+c[/itex]

[itex]\left(ln\sqrt{1+u}-ln\sqrt{1-u}\right)[/itex]=[itex]\frac{1}{\sqrt{kmg}}[/itex][itex]\left(\frac{t}{m}+c\right)[/itex]

e[itex]^{ln\sqrt{1+u}}[/itex]e[itex]^{-ln\sqrt{1-u}}[/itex]=ce[itex]^{\frac{t\sqrt{kmg}}{m}}[/itex]

e[itex]^{ln\sqrt{1+\sqrt{\frac{k}{mg}}v}}[/itex]e[itex]^{-ln\sqrt{1-\sqrt{\frac{k}{mg}}v}}[/itex]=ce[itex]^{\frac{t\sqrt{kmg}}{m}}[/itex]

sunnyceej

With the initial condition: v(0)=0, c=1 and
e[itex]^{ln\sqrt{1+\sqrt{\frac{k}{mg}v}}}[/itex]e[itex]^{-ln\sqrt{1+\sqrt{\frac{k}{mg}v}}}[/itex]=e[itex]^{\frac{t\sqrt{kmg}}{m}}[/itex]

This gives me velocity if I know time, but I know height instead. How do I figure that into my problem?

SammyS

OK.

Let's go back & start from scratch.

You have:

[itex]\displaystyle m(dv/dt)=mg-Kv^2\,.[/itex]

If v is regarded to be a function of x rather than a function of t, use the chain rule.

[itex]\displaystyle \frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=v\frac{dv}{dx}[/itex]

So your differential equation becomes:

[itex]\displaystyle mv\frac{dv}{dx}=mg-Kv^2\,.[/itex]

After separation of variables, this looks easier to integrate than the initial differential equation.

sunnyceej

Thanks. That is a lot easier to integrate. After integrating, I get

v²=[itex]\frac{mg}{k}[/itex]-ce[itex]^{\frac{2kx}{m}}[/itex]

with initial value V(0)=0, I get c=[itex]\frac{mg}{k}[/itex]

so v=[itex]\sqrt{\frac{mg}{k}\left(1-e^{\frac{2kx}{m}}\right)}[/itex]

x would have to represent the height fallen, and it will have to be negative, otherwise I get an imaginary number.

Is this right?

sunnyceej

Thanks for all your help SammyS!

SammyS

Sorry, I didn't get to this yesterday !

I got a similar, but a little different, result.

[itex]\displaystyle v^2=\frac{m}{K}\left(g-Ce^{-(2K/m)x}\right)\,,[/itex] with C = g .

This is the same as what you got except for the sign of the exponent.

sunnyceej

I reworked the problem and got a negative exponent as well.
Thanks again for all your help!