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Velocity of falling object from different heights using DE

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sunnyceej
#1
Feb27-12, 10:49 PM
P: 14
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!
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SammyS
#2
Feb27-12, 11:38 PM
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Quote Quote by sunnyceej View Post
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!
Of course it's separable.

m(dv/dt)=mg-Kv2 → [itex]\displaystyle \frac{dv}{mg-Kv^2}=\frac{dt}{m}\,.[/itex]
sunnyceej
#3
Feb28-12, 12:50 AM
P: 14
Quote Quote by SammyS View Post
Of course it's separable.

m(dv/dt)=mg-Kv2 → [itex]\displaystyle \frac{dv}{mg-Kv^2}=\frac{dt}{m}\,.[/itex]
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=cet/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=180et/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
#4
Feb28-12, 05:17 PM
P: 14
Velocity of falling object from different heights using DE

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
#5
Feb28-12, 07:33 PM
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Quote Quote by sunnyceej View Post
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=cet/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=180et/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:
[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
#6
Feb28-12, 09:58 PM
P: 14
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
#7
Feb28-12, 11:46 PM
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Quote Quote by sunnyceej View Post
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
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
#8
Feb29-12, 11:12 AM
P: 14
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
#9
Mar2-12, 11:57 AM
P: 14
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
#10
Mar2-12, 12:22 PM
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Quote Quote by sunnyceej View Post
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?
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
#11
Mar3-12, 03:16 PM
P: 14
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
#12
Mar4-12, 01:46 PM
P: 14
Thanks for all your help SammyS!
SammyS
#13
Mar4-12, 05:00 PM
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Quote Quote by sunnyceej View Post
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?
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
#14
Mar6-12, 06:36 PM
P: 14
I reworked the problem and got a negative exponent as well.
Thanks again for all your help!


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