Parachutist falling with quadratic drag

In summary: I now have a function that relates V to x and I have to integrate that to get x = f(t). Then I plug in t = 180 seconds and I get the answer. So after the substitution I now have:x = Vt((arctanh(V/Vt)) - (arctanh(Vo/Vt)) = (Vt^2)ln(cosh(Vt(g(t-to))/Vt) - (g(Vt^2)ln(cosh(Vo))) + (Vt^2)(tanh(Vt(g(t-to))/Vt)) + (g(Vt^2)t + (g(Vt^2)to)I'm not sure if the last two terms should be multiplied by
  • #1
Jukai
13
0
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..
 
Last edited:
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  • #2
Jukai said:

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! :approve:. 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! :smile:
 
  • #3
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.
 
Last edited:

Related to Parachutist falling with quadratic drag

1. How does quadratic drag affect a parachutist's fall?

Quadratic drag is a type of air resistance that increases as an object's velocity increases. This means that as a parachutist falls faster, the air resistance they experience also increases. As a result, quadratic drag can significantly slow down the parachutist's fall.

2. What is the equation for calculating the force of quadratic drag?

The equation for quadratic drag force is F = 1/2 * p * v^2 * Cd * A, where F is the drag force, p is the density of the fluid (in this case, air), v is the velocity of the object, Cd is the drag coefficient, and A is the cross-sectional area of the object.

3. How does the shape of a parachute affect the amount of quadratic drag?

The shape of a parachute can greatly impact the amount of quadratic drag it experiences. A larger surface area and a more streamlined shape can increase the amount of drag, while a smaller surface area and a more open shape can decrease it. This is why parachutes are designed with a specific shape and size to control the amount of drag for safe and controlled descent.

4. Is quadratic drag the only force acting on a parachutist during a fall?

No, quadratic drag is not the only force acting on a parachutist. The force of gravity is also acting on the parachutist, pulling them towards the ground. Additionally, there may be other forces at play depending on external factors such as wind or turbulence.

5. How does altitude affect the amount of quadratic drag experienced by a parachutist?

The higher the altitude, the less dense the air is. This means that at higher altitudes, a parachutist will experience less quadratic drag compared to lower altitudes. As a result, a parachutist may fall faster at higher altitudes, but they will also experience less resistance from the air.

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