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Velocity of falling object from different heights using DE 
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#1
Feb2712, 10:49 PM

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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=mgKv^{2}, 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)=mgKv^{2}, since mg = 180 and I know terminal velocity, I converted 120mph to fps and got k=1.614E6. Using g=32fps^{2}, I got m=5.6. Then dividing by m, I get dv/dx=32 1.614E6v^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! 


#2
Feb2712, 11:38 PM

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


#3
Feb2812, 12:50 AM

P: 14

My other thought is to go the natural log route: ln(mgkv^{2})=t/m+c then raising both sides to e: mgkv^{2}=ce^{t/m} using v(0)=0 as my initial condition, c=180, and plugging in other values (mentioned in first post) gives me 1801.614E^{6}v^{2}=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)? 


#4
Feb2812, 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. 


#5
Feb2812, 07:33 PM

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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}{mgKv^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}{mgKv^2}= \sqrt{\frac{1}{Kmg}}\ \int\frac{du}{1u^2}\,.[/itex]Notice that [itex]\displaystyle\frac{1}{1u^2}=\frac{1}{2(1+u)}+\frac{1}{2(1u)}\ .[/itex] 


#6
Feb2812, 09:58 PM

P: 14

So this gives me
[itex]\int\frac{dv}{mgkv^{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}{1u}[/itex]) = [itex]\int\frac{dt}{m}[/itex] becomes [itex]\sqrt{\frac{k}{mg}v}(\frac{1}{2}[/itex]ln[itex]\left1+u\right[/itex] + [itex]\frac{1}{2}[/itex]ln[itex]\left1u\right[/itex]) = [itex]\frac{t}{m}[/itex]+c then [itex]\sqrt{\frac{k}{mg}v}(ln\sqrt{1+u}[/itex]+ln[itex]\sqrt{1u}[/itex])= [itex]\frac{t}{m}[/itex]+c [itex]\sqrt{\frac{k}{mg}v}(ln\sqrt{1u^{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? 


#7
Feb2812, 11:46 PM

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[itex]\displaystyle\sqrt{\frac{k}{mg}v}(\frac{1}{2}[/itex]ln[itex]\left1+u\right[/itex] − [itex]\frac{1}{2}[/itex]ln[itex]\left1u\right[/itex]) = [itex]\frac{t}{m}[/itex]+c 


#8
Feb2912, 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{1u}\right)[/itex]=[itex]\frac{t}{m}+c[/itex] [itex]\left(ln\sqrt{1+u}ln\sqrt{1u}\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{1u}}[/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] 


#9
Mar212, 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? 


#10
Mar212, 12:22 PM

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Let's go back & start from scratch. You have: [itex]\displaystyle m(dv/dt)=mgKv^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}=mgKv^2\,.[/itex]After separation of variables, this looks easier to integrate than the initial differential equation. 


#11
Mar312, 03:16 PM

P: 14

Thanks. That is a lot easier to integrate. After integrating, I get
v^{2}=[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(1e^{\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? 


#12
Mar412, 01:46 PM

P: 14

Thanks for all your help SammyS!



#13
Mar412, 05:00 PM

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I got a similar, but a little different, result. [itex]\displaystyle v^2=\frac{m}{K}\left(gCe^{(2K/m)x}\right)\,,[/itex] with C = g . This is the same as what you got except for the sign of the exponent. 


#14
Mar612, 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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