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Diver Time & Velocity Question

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  1. Mar 2, 2016 #1
    1. The problem statement, all variables and given/known data

    Consider the Olympic sport of diving. The more time the athlete spends in the air, the greater the potential to complete difficult maneuvers. Off of a 3m spring board with a stiffness of 5920N/m, a diver with a mass of 52kg will maximally deflect the board 0.5m from it's equilibrium position. Assume that all of the strain energy stored in the boards returned to the diver in the vertical direction.
    a) How much time will the diver be in the air when using the 3m spring board?
    b) What will the diver's velocity be when they hit the water?
    c) How much time will a diver spend in the air when diving off of a 10m platform?
    d) What will the diver's velocity be when they hit the water?

    2. Relevant equations
    t=(2d/g)1/2
    KE=1/2mv2
    SE=1/2kΔx2

    3. The attempt at a solution

    Sorry for putting a photo in, if you can't make it out let me know and I'll type it out!
    IMG_0244.jpg

    Since the velocities ended up being the same I think I may have gone about it the wrong way...
     
  2. jcsd
  3. Mar 2, 2016 #2

    BvU

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    When does your first relevant equation apply ? Can you use it for this exercise ?
    b) and d) look very much the same to me. Would you get the same answer for a 30 m board too :smile: ?
     
  4. Mar 2, 2016 #3
    I figured when the diver jumps they're like a projectile so that's why I used that time equation.
    Yes, if I continued to use that method, but logically I know that that wouldn't be the case I just am unsure how else to go about it.
     
  5. Mar 2, 2016 #4
    <<Moderator note: Response to deleted post removed>>
    So for c) I did this:
    t=(2h/g)1/2
    t=[(2)(9.5)/(9.81)1/2
    t=1.39s

    Used a height of 9.5m instead of 10m because the board was deflected 0.5m.

    Aaaaaaand, this for d)
    v=(2gh)1/2
    v=[2(9.81)(9.5)]1/2
    v=13.65m/s

    Used a height of 9.5m instead of 10m because the board was deflected 0.5m.
     
    Last edited by a moderator: Mar 2, 2016
  6. Mar 2, 2016 #5

    RUber

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    Using initial height of 9.5m doesn't make sense. One, because the platform doesn't deform like the spring board. Two, because that would be assuming that the diver is simply falling off the deformed board without taking advantage of any upward bounce.
    What you calculated for the strain energy from the spring board looks to be correct, I would assume that velocity would be reached when the board returns to the 3m point where max velocity occurs, not at the 2.5m point where velocity is clearly equal to zero.
     
  7. Mar 2, 2016 #6
    I didn't notice the change in wording, thanks for pointing it out!
     
  8. Mar 2, 2016 #7
    As a former teacher of H.S. physics, I find that the rules of this forum actually prevent what I consider to be "help".
    So having unintentionally violated, and not being in agreement with the one rule that states "no solutions allowed", I will no longer participate.
     
  9. Mar 2, 2016 #8
    Giving a solution to someone is just like giving them a fish but not teaching them how to fish. If you teach your students through solutions only without making them understand the concept by them self, They won't solve most of the questions.

    I am a H.S student and that is what I believe.

    Sorry for going out of the topic. Keep trying to solve OP until you succeed and you will feel the achievement of solving something by your own
     
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