Solving ODE Problem: Air Resistance Proportional to Velocity Squared

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SUMMARY

The forum discussion addresses solving a second-order ordinary differential equation (ODE) related to an object falling under the influence of gravity and air resistance proportional to the square of its velocity. The equation derived is m(d²y/dt²) = mg - kv², where k is a constant related to air resistance. The solution involves integrating the equation to express velocity as a function of position, v(y), and subsequently integrating to find the position as a function of time, y(t). The discussion highlights the use of integration techniques and variable substitution to arrive at the final solution.

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  • Study the derivation of second-order ODEs in physics, particularly those involving forces like air resistance
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Students studying physics or engineering, mathematicians interested in differential equations, and anyone working on modeling motion under the influence of forces such as gravity and air resistance.

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[SOLVED] ODE problem

Homework Statement


A situation in which the air resistance is proportional to the velocity of an object squared.

Object dropped off of a building with height 100m.

F = -mg + Fr Fr=.5*cw*p*A*v^2

cw=.5 p = air density = 1?



Homework Equations



I need to come about with a solution for y(t)

The Attempt at a Solution



I've used Mathcad to help me with a problem when the velocity is proportional to only the velocity linearly, using equations derived analytically.

I just don't know where to begin with this one. I can't derive it analytically to obtain an expression for v(t) so that I can throw that into my runga-kutta solver along with the force expression as the two time derivatives.

I have tried using ODE solver, but it fails to produce anything. It works for two iterations, up to t=2, then quits on me when I try to obtain more results.

I believe my best approach is to solve this analytically, but I can't do it, my prof. did something strange with letting the variable be v^2, but the exercize he did this in was a few weeks ago, and foggy in my mind. Thanks for any help!
 
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From Newton's 2nd Law

m\,\frac{d^2\,y}{d\,t^2}=m\,g-k\,v^2\Rightarrow \frac{d^2\,y}{d\,t^2}=g-\frac{k}{m}\,v^2 \quad (1)

where k=constant. You can write the accerelation as

\frac{d^2\,y}{d\,t^2}=\frac{d\,v}{d\,t}=\frac{d\,y}{d\,t}\frac{d\,v}{d\,y}=v\,\frac{d\,v}{d\,y}

thus (1) reads

v\,\frac{d\,v}{d\,y}=g-\frac{k}{m}\,v^2\Rightarrow \frac{v\,d\,v}{g-\frac{k}{m}\,v^2}=d\,y

Integrate the above equation to obtain v=v(y). Then integrate
v=\frac{d\,y}{d\,t}
to obtain y(t).
 
ahh... thank you very much. Didn't think about the change of base.
 
Integrating
\frac{v\,d\,v}{g-\frac{k}{m}\,v^2}=d\,y
yields

v=-\sqrt{-\frac{m\,g}{k}+c_1\,e^{-2k\,y/m}}\Rightarrow \int\frac{d\,y}{\sqrt{-\frac{m\,g}{k}+c_1\,e^{-2k\,y/m}}}=-\int d\,t

For the first integral make the change of variables

\sqrt{-\frac{m\,g}{k}+c_1\,e^{-2k\,y/m}}=z,\, d\,y=-\frac{z}{g+\frac{k}{m}\,z}\,d\,z
 
Last edited:
you are a lifesaver. Thanks a ton.
 

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