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Skydiving problem

  1. Oct 7, 2006 #1
    When the parachute is fully open, the effective drag coefficient of the sky diver plus parachute increases to 60.0 kg/m. What is the drag force F_drag acting on the sky diver immediately after she has opened the parachute?

    i know that F_drag = Dv^2

    here D is given which is 60 kg/m. and i found v by the equation v= square root of ((m.g)/D)

    so i got 13.06 for V^2
    and then i put the values in the equation F_drag = D.V^2
    = 783. 6

    is this right?
     
  2. jcsd
  3. Oct 7, 2006 #2
    and what about V_termainal?

    V_terminal = square root of ((m.g)/D)
    right?
     
  4. Oct 7, 2006 #3

    arildno

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    This is totally wrong!
    If the parachute opens when a person has reached the terminal velocity gained when NOT having the parachute open, then you must use the D-value valid prior to opening the parachute. That D-value is NOT 60kg/m, since it is explicitly stated that that is the value of D AFTER the parachute has opened.
     
  5. Oct 7, 2006 #4
    Aridno i am not getting you. can you be more clear about D's value? and what equation and values should i use? and i ve another question about V_terminal..i ve posted it. can you look at it?
     
  6. Oct 7, 2006 #5

    arildno

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    Well, if you haven't got any other D-value, or haven't got any velocity value prior to opening the parachute, you can't answer the question.

    By opening the parachute, you are changing the physical parameters.
    If you are to calculate the air resistance force now, you NEED to know what the velocity at that time is!!

    This can be done in basically two ways:
    1. Either that velocity is given, for example that the parachute is opened when at rest.

    2. Or you know that terminal velocity has been reached prior to opening the parachute, in which case you may calculate the terminal velocity by using the D-value valid prior to opening the parachute.
     
    Last edited: Oct 7, 2006
  7. Oct 7, 2006 #6
    A sky diver of mass 80.0 kg (including parachute) jumps off a plane and begins her descent.

    Throughout this problem use 9.80 m/s^2 for the magnitude of the acceleration due to gravity.


    this is the information given..
     
  8. Oct 7, 2006 #7

    arildno

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    In that case, this being free fall, you must know either the height he has fallen prior to opening the parachute, or the time gone by to solve it.

    Since this has not been given, we must see if there possibly is a TACIT assumption made in the exercise, that enables us to solve it anyway.

    The only tacit assumption of any merit is that he opens the parachute IMMEDIATELY when he jumped; that is, his vertical velocity at the time of opening the parachute is ZERO.
    However, even this is insufficient to solve the problem, since relative to the air, the jumper has a HORIZONTAL velocity equal to that of the plane!

    Thus, you must at the very least know the plane's velocity in order to solve the problem in this manner!
     
  9. Oct 7, 2006 #8
    Terminal Velocity is about 90-120mph i think, incase if you wanted to estimate from that. 120mph if your in Tandem or a fat & dense dude.
     
  10. Oct 7, 2006 #9
    how it is 90-120 mph?
     
  11. Oct 7, 2006 #10
    A sky diver of mass 80.0 kg (including parachute) jumps off a plane and begins her descent.

    Throughout this problem use 9.80 m/s^2 for the magnitude of the acceleration due to gravity.


    What is the terminal speed v_terminal of the sky diver when the parachute is opened?
     
  12. Oct 7, 2006 #11
    Well, u got the right answer for v_terminal anyway in ur 1st post. (v^2 = 13.06)

    U still need more info about before the parachute was opened to get the drag just after.
    The drag force u calculated (783. 6) is the drag force when the skydiver achieved v_terminal, which is a little while after the parachute opens.

    Y dont u write out d whole question in 1 post.
     
  13. Oct 7, 2006 #12
    Sorry, Im talking about actual free-fall terminal velocity of a skydiver not the terminal velocity of a parachutist. That figure is not one i calculated myself, but one reported to me by my skydiving instructor. I was just trying to fill in the missing info.

    The force exherted on the diver is a matter of the deceleration felt when the chute opens, the information missing from your original post ( i must admit i only scanned ) was with regards to the free-fall velocity. Thats why i said you could estimate it.

    Of course, the force is also effected by the time it takes for the chute to open, but i assume thats being neglected for simplicity.

    To get the rate of change in velocity you need to know the free-fall terminal velocity, the parachute deployed terminal velocity and the time for deployment. Sorry if i wasnt very clear.. :confused:
     
    Last edited: Oct 7, 2006
  14. Oct 8, 2006 #13
    hey, Blackwizard. this is the entire question ok..
    Information Given:
    A sky diver of mass 80.0 kg (including parachute) jumps off a plane and begins her descent.

    Throughout this problem use 9.80 m/s^2 for the magnitude of the acceleration due to gravity.

    question 1:

    What is the terminal speed v_terminal of the sky diver when the parachute is opened?


    question 2:

    When the parachute is fully open, the effective drag coefficient of the sky diver plus parachute increases to 60.0 kg/m. What is the drag force F_drag acting on the sky diver immediately after she has opened the parachute?

    question 3:

    For an object falling through air at a high speed v, the drag force acting on it due to air resistance can be expressed as

    F=Kv^2,

    where the coefficient K depends on the shape and size of the falling object and on the density of air. For a human body, the numerical value for K is about 0.250 kg/m.

    Using this value for K, what is the terminal speed v_terminal of the sky diver?


    question 4:

    At some point during her free fall, the sky diver reaches her terminal speed. What is the magnitude of the drag force F_drag due to air resistance that acts on the sky diver when she has reached terminal speed?
     
  15. Oct 8, 2006 #14
    Anyone? please help..
     
  16. Oct 8, 2006 #15
    Erm, ill attemp t it

    To summarise:
    Variables:
    [tex]m=80kg[/tex]
    [tex]g=9.8m\ s^{-2}[/tex]
    [tex]k_{f}=0.25kg\ m^{-1}[/tex]
    [tex]k_{p}=60kg\ m^{-1}[/tex]
    [tex]V_{f}=?[/tex] (Terminal Velocity in Free-Fall Q3)
    [tex]V_{p}=?[/tex] (Terminal Velocity under parchute canopy Q1)

    Formula:
    [tex]F_{d}=kV^{2}[/tex]

    [tex]F_{w}=mg[/tex]

    At terminal Velocity:
    [tex]F_{d}=F_{w}[/tex]

    [tex]kV^{2}=mg[/tex]

    Rearrange for [tex]V_{f}[/tex]:
    [tex]V_{f}\ = \ \sqrt{\frac{mg}{k_{f}}}[/tex]

    [tex]V_{f} = \sqrt{\frac{784}{0.25}} = 2\times28 = 56m\ s^{-2}[/tex]

    [tex] 56m\ s^{-2} = 201.6km\ h^{-1} = 126mph[/tex]
     
    Last edited: Oct 8, 2006
  17. Oct 8, 2006 #16
    so for question 3, ans is 56 m/s. answer should be in m/s units.

    and i am still not getting other questions. can you help me out with it?
     
  18. Oct 8, 2006 #17
    Well, it follows that Q1 is the same method as Q3, but the coefficient of drag has changed for an open chute. So plug in the value and see what ya get (tbh, it seemed too low to me).

    The drag force equation is there, just put in the values.
     
  19. Oct 9, 2006 #18
    well its not the right answer. and what i should do now? can you please give me some more clarification how you did it? and some more suggestions...
     
  20. Oct 9, 2006 #19
    Are you sure thats wrong?

    I mean, its seems right to me. No net acceleration at terminal velocity implies that mg = kV^2. The numbers were all neat like your typical book example, even the roots worked out easy to calculate.....seemed good to me.

    What is the answer your given for Q3?
     
  21. Oct 9, 2006 #20
    hey man can you help me solving Q1, Q2 and Q4?
     
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