Parachutist falling with quadratic drag

[Jukai]

MAJOR EDIT: I fixed my position integral and got the answer to 1.ii)

Homework Statement

A parachutist jumps from a helicopter that's not moving. The mass of the person is 80kg, the quadratic drag is f = -Cv^2 where C = 3,52. Neglect the Earth's rotation effect. The parachute is opened as soon as the person jumps.

i) Is there a terminal speed? if yes, what is it?

ii) What is the height at which the helicopter must be so that the parachutist stays in the air exactly 3 minutes if her initial speed is 0.

iii) If the parachutist falls with a constant speed (the terminal speed), at what height must the helicopter be so that the person stays in the air exactly 3 minutes.

Homework Equations

ma = mg -Cv^2 (1)

Vt= (mg/C)^(1/2) where Vt : terminal speed

[tex]\int(1/(1 - ((x^2)/(a^2))) = a(arctanh(x/a))[/tex]
[tex]\int((tanh(x)dx)= ln(cosh(x)))[/tex]

tanh(x) = (e^x - e^-x)/(e^x + e^-x)
cosh(x) = (e^x + e^-x)/2

The Attempt at a Solution

i) I got this one right. I made the logical argument that if the parachutist reaches a terminal speed, there is no more acceleration, therefore (1) becomes
mg = CV^2 and Vt = 14,92 m/s

The next questions is where I'm having intense difficulty

ii) I divide by m everywhere in (1) and factor g from the right side and I get this:

dv/(1 - ((V^2)/(Vt^2))) = gdt

Integration gives me the following:

Vt((arctanh(V/Vt)) - (arctanh(Vo/Vt)) = g(t - to)

where to and Vo are the initial time and speed

Since I'm looking for height, I need to integrate V once more. I isolate V as follows and integrate as follows:

V = Vt( (tanh(g(t - to)/Vt)) + (arctanh(Vo/Vt)))

X - Xo = ((Vt^2)/g)(ln(cosh( (g(t-to)/Vt) + (arctanh(Vo/Vt)))))

Now when I put t = 180, it works.

iii) I'm not even sure how I'm supposed to integrate when V is constant throughout the fall...

For this question, my initial speed is Vt and the problem with arctanh(1) going to infinity persists..

 

[collinsmark]

Homework Equations

ma = mg -Cv^2 (1)

Vt= (mg/C)^(1/2) where Vt : terminal speed

[tex]\int(1/(1 - ((x^2)/(a^2))) = a(arctanh(x/a))[/tex]
[tex]\int((tanh(x)dx)= ln(cosh(x)))[/tex]

tanh(x) = (e^x - e^-x)/(e^x + e^-x)
cosh(x) = (e^x + e^-x)/2

The Attempt at a Solution

i) I got this one right. I made the logical argument that if the parachutist reaches a terminal speed, there is no more acceleration, therefore (1) becomes
mg = CV^2 and Vt = 14,92 m/s

The next questions is where I'm having intense difficulty

ii) I divide by m everywhere in (1) and factor g from the right side and I get this:

dv/(1 - ((V^2)/(Vt^2))) = gdt

Integration gives me the following:

Vt((arctanh(V/Vt)) - (arctanh(Vo/Vt)) = g(t - to)

where to and Vo are the initial time and speed

Since I'm looking for height, I need to integrate V once more. I isolate V as follows and integrate as follows:

V = Vt( (tanh(g(t - to)/Vt)) + (arctanh(Vo/Vt)))

Great job so far! . But you can simplify things quite a bit by defining v₀ = 0 and t₀ = 0. You don't really lose any generality by doing so for this particular problem, and it gets rid of that second term.

X - Xo = Vt(ln(cosh( (g(t-to)/Vt) + (arctanh(Vo/Vt)))))

when I put t = 3 minutes = 180 seconds, I don't get the right answer.

Try this on for size. You already know that

[tex] \int \tanh x \ dx= \ln (\cosh x) + K[/tex]

where K is an arbitrary constant.

Substituting x = ay, where a is a constant, and noting that dx = a dy --> dy = (1/a)dx,

[tex] \int \tanh (ay) \ dy= \frac{\ln (\cosh ay)}{a} + K[/tex]

Good luck!

 

[Jukai]

Thank you very much =), it's too easy to forget the substitution trick for integration..

last edit: (for those who care =) )So I found out iii), I just had to integrate to x starting from ma= -CV + mg where the right side are constants.