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Felix Baumgartner's parachute jump from space -- forces and velocities...

  1. Jan 24, 2016 #1
    1. The problem statement, all variables and given/known data
    IMG_20160124_143418.jpg
    a)ii) Calculate the upward force F acting on Baumgartner at this point. (3 marks)
    total mass of Baumgartner = 95 kg
    g = 9.8 ms-2
    b) Describe the shape of the graph between 30s and 70s. Explain the velocity changes in terms of changes in the air through which Baumgartner was falling. (3 marks)
    c) It has been claimed that Baumgartner fell more than 35km in the 260 seconds before he opened the parachute. Use the graph to check whether this is correct. Show your method clearly. (3 marks)

    2. Relevant equations
    3. The attempt at a solution

    Mostly just looking to check my answers and understand how an exam board would want c) answered.

    a)i) U at t=20 is 200m/s.
    V at t=40 is 340m/s.
    (340 - 200) / 20 = 7 ms-2, QED.
    a)ii) Hmm. Usually I'd do 95 * 9.8 = 931 N. But his acceleration at the point was around 7ms-2, so they could be trying to catch me out by giving me g here. I don't understand what the force would be contingent on. Doesn't free fall imply constant acceleration from g?
    b) The shape is a negative parabola. Terminal velocity is reached at t ~~ 45, at which point he begins to decelerate due to increased atmospheric density and hence increased air resistance.
    c) From t=0 to t=50 the curve roughly forms a triangle with the x-axis. 0.5*350*50 = 8750m.
    From t=50 to t=150 a rough trapezium is formed. 100 * (350+80) / 2 = 21500m.
    From t=150 to t=260 another rough trapezium is formed. 110 * (80+50) / 2 = 7150m.
    8750 + 21500 + 7150 = 37400m = 37.4 km. The claim is correct. (Is this proper exam technique?)
     
    Last edited: Jan 24, 2016
  2. jcsd
  3. Jan 24, 2016 #2

    CWatters

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    Clearly not or his acceleration would be 9.8 not 7m/s2.

    Make a free body diagram showing all the forces acting on him.

    Remember that in the equation F=ma...

    "F" is the net (or total) force acting on him (eg not just that due to gravity)
    "a" is his actual acceleration which in this case isn't "g"
     
  4. Jan 24, 2016 #3
    Right, sometimes the obvious answer is the right one. 95 * 7 = 665N.
    So questions that ask you something like this where you end up calculating mg for upwards force are neglecting air resistance.
     
  5. Jan 24, 2016 #4

    CWatters

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    I think I would want a bit more explanation than that. I think I would write a comment on the air density and air resistance at three points (eg 30S, 45S and 70S). Hint: My answer would include the words "less than", "equal to" and "greater than"

    It asks you to explain or show your method clearly. I'm not sure I would accept "the curve forms a triangle" as an explanation of your method. I can see what you are doing when you write "0.5*350*50" but I think you should explain it in English to get the full marks.
     
  6. Jan 24, 2016 #5

    CWatters

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    You should read the question carefully :-) The question asks about the "upward force F acting on Baumgartner".
     
  7. Jan 24, 2016 #6
    Right. At 30s drag from air resistance is less than the force of weight, so Baumgartner is accelerating.
    At 45s drag = force of weight, terminal velocity.
    At 70s drag is greater than force of weight, hence deceleration.
    Not sure if that's exactly the right terminology.

    Area under graph = displacement (ms-1 * s).

    Force is a vector, so -655N? If not I don't have an answer that still abides by F = ma.
     
  8. Jan 24, 2016 #7

    CWatters

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    That's better however..
    No that's wrong.

    Ok so lets recap...

    The downward force due to gravity is 931N.
    You say the upward force due to air resistance is either 655N or -655N.

    As a check, try calculating the net force acting on him and the resulting acceleration.
     
  9. Jan 24, 2016 #8
    R.F/95 = 7ms-2. 95 * 7 = 655, so this was my resultant force downwards.
    931 - 655 = 276N upwards?
     
  10. Jan 24, 2016 #9

    CWatters

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    Correct. The upward force due to air resistance is 276N.
     
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