Velocity of falling object from different heights using DE

[sunnyceej]

Homework Statement

Find the velocity of a 180 lb object falling from a state of rest from 5 ft, 15 ft, 25 ft, and 45 ft

Homework 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.

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]

Homework Statement

Find the velocity of a 180 lb object falling from a state of rest from 5 ft, 15 ft, 25 ft, and 45 ft

Homework 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.

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!

Of course it's separable.

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

 

[sunnyceej]

Of course it's separable.

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

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]

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)?

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:

\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)}​

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

The integral then becomes:

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

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

 

[sunnyceej]

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

becomes

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

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

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

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

Is this right?

 

[SammyS]

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

becomes

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

That should be a minus sign.

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

 

[sunnyceej]

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

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

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

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

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

 

[sunnyceej]

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

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

 

[SammyS]

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

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

OK.

Let's go back & start from scratch.

You have:

\displaystyle m(dv/dt)=mg-Kv^2\,.​

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

\displaystyle \frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=v\frac{dv}{dx}

So your differential equation becomes:

\displaystyle mv\frac{dv}{dx}=mg-Kv^2\,.​

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=\frac{mg}{k}-ce^{\frac{2kx}{m}}

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

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

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]

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

v²=\frac{mg}{k}-ce^{\frac{2kx}{m}}

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

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

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

Is this right?

Sorry, I didn't get to this yesterday !

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

\displaystyle v^2=\frac{m}{K}\left(g-Ce^{-(2K/m)x}\right)\,, 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!