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

In summary, the motion of a parachuter involves key concepts such as terminal velocity, time of flight, and distance traveled. Terminal velocity is the maximum speed reached when the force of gravity is balanced by air resistance, resulting in no further acceleration. The time of flight refers to the duration from jump to landing, which is influenced by factors like altitude and drag. The distance covered during the fall depends on the initial drop height and the effects of air resistance, leading to a gradual increase in speed until terminal velocity is achieved. Understanding these factors is crucial for analyzing parachuting dynamics.
  • #1
I_Try_Math
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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.
 
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  • #2
I_Try_Math said:
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.
 
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  • #3
haruspex said:
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.
 
  • #4
I_Try_Math said:
Oh yes I meant it to be mg instead of ma.
I edited my post to reflect that.
 
  • #5
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?
 
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  • #6
haruspex said:
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.

1706331853214.png


Here is the exact text of the question with the given formula for F_D.
 
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  • #7
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.
 
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  • #8
Or that the book has a typo in the answer key...
 
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  • #9
Delta2 said:
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.
 
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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?
 
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FAQ: Motion of a Parachuter (Terminal Velocity, Time of Flight, Distance)

What is terminal velocity and how is it achieved?

Terminal velocity is the constant speed that a freely falling object eventually reaches when the resistance of the medium through which it is falling prevents further acceleration. For a parachuter, this occurs when the force of gravity pulling them downwards is balanced by the air resistance pushing upwards, resulting in zero net acceleration.

How long does it take for a parachuter to reach terminal velocity?

The time it takes to reach terminal velocity depends on several factors, including the parachuter's body position, weight, and the altitude from which they jump. Generally, a skydiver in a belly-to-earth position reaches terminal velocity in about 10-15 seconds, covering approximately 1,000-1,500 feet in that time.

How does the deployment of a parachute affect terminal velocity?

When a parachute is deployed, it significantly increases the surface area and thus the air resistance. This drastically reduces the terminal velocity from around 120 mph (for a free-falling skydiver) to about 15-25 mph, allowing for a safe and controlled descent to the ground.

What factors influence the time of flight for a parachuter?

The time of flight for a parachuter is influenced by the altitude of the jump, the time taken to reach terminal velocity, and the time spent descending under the parachute. Weather conditions, such as wind speed and direction, can also affect the descent time.

How can you calculate the distance fallen before reaching terminal velocity?

The distance fallen before reaching terminal velocity can be estimated using the equations of motion under constant acceleration (gravity) and taking into account the increasing air resistance. A simplified approach involves using the average speed during the acceleration phase and the time taken to reach terminal velocity. More precise calculations would require solving differential equations that account for changing velocity and air resistance.

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