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Skydiver falling (i am stuck)

  1. Dec 18, 2006 #1
    1. The problem statement, all variables and given/known data
    "An 80.0 kg skydiver (initially at rest) jumps out or an airplane at an altitude of 1000.0 m and opens the parachute at some altitude. Assuming that the totalretarding force on the skydiver is constant at 50.0 N with the parachute closed and the constant at 3600.0 N with the parachute open, at what altitude should the skydiver open the parachute so that he lands with a velocity of 5.00 m/s on the ground?"

    Quite a mouth full.

    2. Relevant equations

    Vf^2 = Vi^2 + 2ax
    Sigma F= MA

    3. The attempt at a solution

    Okay im really kinda stuck on this one.
    I tried figuring out the acceleration by doing:

    784.8 N -3600N = 80 kg(a)
    but that yields an acceleration of -35 m/s^2, And that can't be right.

    I also drew a free body diagram but i can't post that up.

    I know that the correct answer is 207 m because the teacher gave that to us.

    Can anyone help me?
  2. jcsd
  3. Dec 18, 2006 #2

    Doc Al

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    Staff: Mentor

    Why not?

    But that's just the acceleration during the second part of the motion (after the chute opens). What about the first part of the motion?
  4. Dec 18, 2006 #3
    Well i thought of that but i didn't seem to get me anywhere.
    So then if i do do this:
    784.5 N - 50 N = 80 kg (a)
    then i get an acceleration of :9.185 m/s^2

    but where would i go with that?
  5. Dec 18, 2006 #4
    Find the acceleration for each part of the motion (youre on the right track). Then you need to find the right distance for each part of the motion knowing that the total distance is 1000.0m
  6. Dec 18, 2006 #5

    Doc Al

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    Staff: Mentor

    Set up equations for the two parts of the motion and see what you can figure out. You know the initial and final speed, and the total distance.
  7. Dec 18, 2006 #6
    ok so then if i use my equation ^^^.

    and do 5m/s^2= 0 + 2(-35.2 m/s^2)(x)
    i get x = to -0.355 m

    and 5m/s^2 = 0 + 2(9.2m/s^2)(x)
    i get 1.36 m

    and that can't be right. What am i doing wrong?

    Oh and to the website designers it would be a nice plus to have a exponent button.
  8. Dec 18, 2006 #7
    It starts at rest, accelerates positively to a middle velocity, and then decelerates back to 5m/s. Eliminating should get rid of the intermediate velocity, though
  9. Dec 18, 2006 #8
    I am not really helped out by your response could you give me an example? or a clue
  10. Dec 18, 2006 #9


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    Staff Emeritus
    Science Advisor

    There is. If you want to write m/s2 you write m/s[ sup ]2[/ sup ] (without the spaces in the square brackets.)
  11. Dec 18, 2006 #10
    Your top equation should start off (5m/s)^2 - vMIDDLE(^2)
    Your bottom equation should start off (vMIDDLE)^2 - (0m/s)^2

    And make sure you dont use the same distance in both equations
  12. Dec 18, 2006 #11
    yes but that would give me:
    5m/s2 - vmiddle2 = 0 + 2(-35.2 m/s2)(x)

    which gives me to variables to figure out. And i am only in honors physics in 11th grade so I don't know how to do that.
  13. Dec 18, 2006 #12
    You need as many equations as you have unknowns. I count 3 equations with 3 unknowns. You dont know d1, d2, or vmiddle. But you can eliminate vmiddle and easily write d2 in terms of d1 because they sum up to 1000 (think of an angle and its complement). After eliminating, you should be able to substitute and get either d1 or d2. Then you just subtract that from 1000
  14. Dec 18, 2006 #13
    you know what you are saying is slapping on the bell but not ringing it.
    My parents want me to get some sleep can you just set up the equation for me, cause i am still lost. Like what do i do to find the d2?
  15. Dec 19, 2006 #14
    have you learnt energy yet? this problem could be very simply solved by energy.

    just look at the work done by each force. Assuming that the person doesn't go back upward, you can set up an equation.
  16. Dec 19, 2006 #15
    oh i completely forgot about on of the equations that the teacher gave us:
    (sigma)work = 1/2mvf2-1/2mvi2

    I think that the right side of the equation cancels out because there is no vi. But how do i do the work part?
  17. Dec 20, 2006 #16
    well, so you have the work.

    and Work=Force * Distance... Now, assuming that the first net force acts over a distance of x, what would the work done by the first and second force be (in terms of x) (be very careful about the sign of the work)? together, what is the equation that can solve the problem?
    Last edited: Dec 20, 2006
  18. Dec 20, 2006 #17
    thank you i will try it
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