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Ski Jump Ramp question

  1. Oct 19, 2015 #1
    1. The problem statement, all variables and given/known data
    A 72 kg skier leaves the end of a ski-jump ramp with a velocity of 24 m/s directed 25◦ above the horizontal. The skier returns to the ground at a point that is 14 m below the end of the ramp with a total speed of 22 m/s.

    (a) What is the rate at which the gravitational force does work on the skier at the moment when he leaves
    the ramp?

    (b) With what total speed would he have landed if there were no air drag?

    (c) What was the work done on the skier by the air drag?



    2. Relevant equations

    This is what I am uncertain about.

    3. The attempt at a solution

    I think I was able to do part (c) by letting 1/2(m)(v2^2-v1^2) + mgh, Got a large number but it seems correct.


    What I am confused about is where to even start the first two questions, I have been looking through my notes but it seems I've misplaced my notes regarding this chapter which is fairly annoying. Could anyone recommend equations that would help resolve parts a and b?
     
  2. jcsd
  3. Oct 19, 2015 #2

    BvU

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    Hello your majesty,

    being uncertain about something is one thing. But you are uncertain about nothing: there is no relevant equation at all in the place where some are needed !

    You know about potential energy from gravity ? something like mgh ? doing work often means changing energy. m and g don't change much, so it'll have to do with the change in h. What is the rate f change of h at the point of leaving the ramp ? :smile:
     
  4. Oct 19, 2015 #3
    m = 72kg
    g = 9.81m/s^2
    h = 14m

    mgh = 9,888.5 Joules right? Would that be the rate of change when the skier leaves the ramp? :)
     
  5. Oct 19, 2015 #4

    BvU

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    No
    1) the 14 m comes later.
    2) a rate has the dimension of something per unit of time. Work is in Joules, rate of doing work is in Watt (Joules/second)

    When leaving the ramp, what is the rate of change of h ?
     
  6. Oct 19, 2015 #5
    Rate of change is the slope or the first differential with respect to h right? That is my understanding of ROC anyway. Is it the rate of change in terms of time?
    like dh/dt?
    Or instead of dh/dt... dQ/dt? (Q = theta = 25 degrees)
     
  7. Oct 19, 2015 #6

    BvU

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    It is in terms of time. It is dh/dt. :smile:

    But: dQ/dt doesn't help (25 degrees is 25 degrees, right?). Make a little drawing for that scary moment. See how high the skier is at that moment and how fast his/her height increases with time .
     
  8. Oct 19, 2015 #7
    I feel so stupid, this is what a week of a common cold can set you back with. I understand that but I can't seem to see the relation to the question! I feel like I'm missing some sort of information here, or something key. It is probably a very easy thing too if its something to do with rate of change.
     
  9. Oct 19, 2015 #8

    BvU

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    Flattery will get you nowhere. If you throw something in the air with a vertical speed of 10 m/s, its height initially changes at a rate of ... 10 m/s !
     
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