Solve ODE for Free Fall with Drag: Find v(t)

[tua96426]

Homework Statement

Need to solve the differential equation which models the velocity of a sky diver.

Homework Equations

dv/dt = -9.8 + 0.0045Av^2 where A is cross sectional area (assume 0.75m^2)
v(0) = 0

The Attempt at a Solution

ok the problem is to find v(t) as a function time. as they teach us any ODE class, I started with separation of vars and on the left i had (dv/0.0045Av^2 - 9.8) and dt on the right,
then integrate both sides. The right hand side just becomes t + y (constant of integration) and the left needs to be simplified first with partial fractions

I said let c =sqrt(0.0045A) and b= sqrt (9.8)
then we have dv/(cv)^2 - b^2 which is now a difference of perfect squares ..
so partial fractions: A/(cv - b) + B(cv+b)
getting common denominator we get Acv + Ab + Bcv - Bb = 1

so all vars together: Ac + Bc = 0 >>> A+B = 0 therefore A = -B

then looking at the constant terms, I get b(A-B) = 1 >>> A = 1/b + B
so equation the two equations for A we get B = -1/2b and A = 1/2b
B = -1.565 and A = 1.565

so now we have Integral of (1.565/cv-b) + (-1.565/cv-b) dv
making a u substitution: u = cv +/- b >> du = cdv >>> dv = du/c
so we have (-1.565/c)ln|cv-b|+ (-1.565.c)ln|cv+b| = t + y
taking 1.565/c common on the left and multiplying by c/1.565 on both sides.. and getting all ln into one, I got
ln |(cv-b)/(cv+b)| = (c/1.565)(t+y)
let x = -c/1565
so doing more of mumbo jumbo, i get
v(t) = (b/c) ((1 + e^x(t+y))/(1-e^x(t+y)))

some how after applying the initial conditions, I got y = 0, but when I plug in 0 for t after getting the unique soln for v(t), I do not get 0 for velocity; rather I end up dividing by 0..

I researched this on wiki and they have the answer in tanh, which my prof said is possible. I was wondering how to get it in that form.. infact if I am able to express it in tanh it becomes easier for me to do the following problems (need to integrate and so on and so forth) .. any help on this would be gr8 peeps!

thanks
- Sudhi

 

[lanedance]

i think it could be the way you set up the DE

the drag force should oppose velocity. As you're working with v^2 will need to define poitsive velocity downwards

so try

v' = g - cAv^2

 

[gabbagabbahey]

Hi Sudhi, welcome to PF!

so doing more of mumbo jumbo, i get
v(t) = (b/c) ((1 + e^x(t+y))/(1-e^x(t+y)))

some how after applying the initial conditions, I got y = 0, but when I plug in 0 for t after getting the unique soln for v(t), I do not get 0 for velocity; rather I end up dividing by 0..

You seem to have made a slight error in your "mumbo jumbo" steps; you should be getting $$v(t)=\frac{b}{c}\left(\frac{1-e^{x(t+y)}}{1+e^{x(t+y)}}\right)$$

Also, you could have very easily determined y just by plugging in t=0 and v=0 into ln |(cv-b)/(cv+b)| = (c/1.565)(t+y).

I researched this on wiki and they have the answer in tanh, which my prof said is possible. I was wondering how to get it in that form.. infact if I am able to express it in tanh it becomes easier for me to do the following problems (need to integrate and so on and so forth) .. any help on this would be gr8 peeps!

Use the definition of hyperbolic tangent:

$$\tanh(u)=\frac{e^u-e^{-u}}{e^u+e^{-u}}=\frac{1-e^{-2u}}{1+e^{-2u}}$$

 

[gabbagabbahey]

i think it could be the way you set up the DE

the drag force should oppose velocity. As you're working with v^2 will need to define poitsive velocity downwards

so try

v' = g - cAv^2

The DE is fine, it assumes downwards corresponds to a negative v(t), and so the acceleration due to gravity is indeed downwards (-9.81) while the drag force is upwards (+.0045Av^2).

 

[lanedance]

woops - my bad, sorry to send you on the wrong track

 

[lanedance]

and as; for mumbo jumbo, i think you mix up a few negative signs early on

will make life easier as well if you write -t (ie put a negative on the t side) before combining the ln's or taking the exponentials - this is currently hidden in your x... i find its usually easiest to make constants positive as a convention

this will give you the equation starght away with something like a.(1-e^(-kt)) in the numerator so velocity starts at zero, accelerates quickly then heads towards a limiting velocity as the drag forces kick in