Motion of a Parachuter (Terminal Velocity, Time of Flight, Distance)

[I_Try_Math]

Homework Statement

A skydiver is at an altitude of 1520 m. After 10.0 seconds of free fall, he opens his parachute and finds that the air resistance, F_D, is given by the formula F_D = -b*v where b is a constant and v is the velocity. If b = .75 and the mass of the skydiver is 82.0 kg, first set up differential equations for the velocity and the position, and then find: (a) the speed of the skydiver when the parachute opens, (b) the distance fallen before the parachute opens, (c) the terminal velocity after the parachute opens (find the limiting velocity), and (d) the time the skydiver is in the air after the parachute opens.

Relevant Equations

F = ma
F_D = -0.75*v

(a) -98 m/s
(b) 490 m

(c)
My understanding is that at terminal velocity the net force in the y direction must be zero.
Therefore:
F_y = ma = 0

Only drag and weight forces act on the skydiver so:

F_D + mg = 0

F_D = -mg

-0.75*v = -82*(-9.8)

v = -1071.5 m/s

The value I get for v appears to be incorrect based on my textbook and common sense. However, in the abstract, the math appears to be logical. Please help me understand what I am missing.

 

[haruspex]

F_D - ma = 0

Is "a" supposed to be an acceleration, or is "ma" supposed to be the force of gravity?
If the second, with no acceleration, the forces add to zero.

 

[I_Try_Math]

Is "a" supposed to be an acceleration, or is "ma" supposed to be the force of gravity?
If the second, with no acceleration, the forces add to zero.

Oh yes I meant it to be mg instead of ma.

 

[I_Try_Math]

Oh yes I meant it to be mg instead of ma.

I edited my post to reflect that.

 

[haruspex]

There ought to be units for b. If none given then I guess you are right to assume kg/s, which gives that huge velocity.
But at those speeds it should be quadratic. Are you sure it's not $bv^2$?

Do you know what the answer is supposed to be?

 

[I_Try_Math]

There ought to be units for b. If none given then I guess you are right to assume kg/s, which gives that huge velocity.
But at those speeds it should be quadratic. Are you sure it's not $bv^2$?

Do you know what the answer is supposed to be?

The textbook claims the answer for (c) is 107 m/s.

Here is the exact text of the question with the given formula for F_D.

 

[Delta2]

your answer is 1071.5 while the book answer is 107 so it seems the one is 10 times the other, it must be some unit conversion issue that has to do with the units of b=0.75.

 

[Delta2]

Or that the book has a typo in the answer key...

 

[I_Try_Math]

Or that the book has a typo in the answer key...

Ok thank you, I believe I've seen at least one other typo in the answer key so that is possible.

 

[Delta2]

Btw is there a clever shortcut for d) or we have to use the differential equation solution y(t), set $$y(t)=1520-\frac{1}{2}g(10)^2=1020 (g=10m/s^2)$$ and solve for t?

 

[haruspex]

Or that the book has a typo in the answer key...

Or that b should be 7.5, or that the author forgot to convert kg to N…
Even then, 107m/s is much too fast with a parachute. Even without one, human terminal velocity in the horizontal position is half that, largely because it is a quadratic formula.
See https://en.wikipedia.org/wiki/Free_fall#Uniform_gravitational_field_with_air_resistance