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Kinematics motion - Dancers pausing at top of their leap

  1. Dec 19, 2013 #1
    Kinematics motion -- Dancers pausing at top of their leap

    1. The problem statement, all variables and given/known data

    Ice skaters, ballet dancers, and basket ball players executing vertical leaps often give the illusion of "hanging" almost motionless near the top of the leap. To see why this is, consider a leap to maximum height, h. Of the total time spent in the air, what fraction is spent in the upper half (i.e., at y > 0.5h)

    2. Relevant equations
    None


    3. The attempt at a solution

    yf = yi + vit + 0.5gt^2

    yf = 0 + vit + 0.5(-9.8ms^-2)t^2
    yf = vit - 4.9ms^-2 t^2

    Should I find the first derivative of yf and determine the local max? And what does upper half implies? The max time take for the object to travel from ground to max height?
     
    Last edited: Dec 19, 2013
  2. jcsd
  3. Dec 19, 2013 #2

    TSny

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    Edited: If h is the maximum height of the jump, you want to find the ratio of the time to travel the upper half (from y = .5h to the max height y = h and back to y = .5h) to the total time to travel from the ground and back to the ground.
     
    Last edited: Dec 19, 2013
  4. Dec 19, 2013 #3
    Capture.JPG

    By finding the quadratic roots of each of the equation, and taking the time taken for the object to travel up from ground to max height,h and from 0.5h to ma height, h, its possible to find the ratio.
    However, I'm unable to reduce the quadratic equations to their respective roots without any numerical values for vi in place.
     
    Last edited: Dec 19, 2013
  5. Dec 19, 2013 #4
    You did not use the fact that h is the maximum height reached when the object is launched with vi.
    You need to express the time as a function of h (and g) only. So eliminate vi first.

    And of course, in the second equation the initial speed is different. Using the same "vi" for both is confusing.
     
  6. Dec 19, 2013 #5
    View attachment 64943

    Is this right?
     
    Last edited: Dec 20, 2013
  7. Dec 19, 2013 #6

    TSny

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    In going from the 2nd to the 3rd line, did you multiply 2(-9.81) correctly?

    Don't forget the equation vf = vi + at
     
  8. Dec 19, 2013 #7
    Capture.JPG
    Corrected the blunder.

    Is it not possible to solve it via quadratic?
     
  9. Dec 19, 2013 #8

    TSny

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    For this problem, there is no need to plug in values for g.

    You have 02 -vi2 = -2gh.

    What do you get if you solve this for vi in terms of g and h?

    -----

    Yes, you can work with the quadratic equation in t, but it is easier to work with the linear equation.
     
  10. Dec 19, 2013 #9
    Why is the rational for not plugging in values for g?

    vi= sqrt[-2gh]
     
  11. Dec 19, 2013 #10

    haruspex

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    You're overcomplicating it. We don't need to consider the whole leap. The descent half will do. The ascent is just a mirror image of that in time.
    You have the distance, the initial speed (0) and the acceleration. h = gt2/2.
    How long does it take to descend half way?

    Btw, dancers might intentionally exaggerate the effect by the way they move their limbs. The downside is that the maximum height their heads reach is less. A footballer is unlikely to do that.
     
  12. Dec 19, 2013 #11

    TSny

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    The expressions are "cleaner" if left in symbols. Leaving the expressions in terms of g will also show that the final answer for this problem does not depend on the value of g. You would get the same answer for someone jumping on the moon as here on the earth.

    g is usually defined as the magnitude of the acceleration of gravity, which makes g a positive number.
    Note that your result vi = Sqrt[-2gh] has the square root of a negative number if g is considered positive. [EDIT: However, if you wanted to go against convention and take g to be a negative number throughout the problem, then your equation for vi is correct.]

    [Starting with 02 - vi2 = -2gh, you should not end up with a negative sign inside the square root when solving for vi.]
     
    Last edited: Dec 19, 2013
  13. Dec 19, 2013 #12
    it takes t = sqrt[2h/g]
    I find it easier to find the ratio between time taken to travel from ground to h and time taken to travel from h/2 to h.

    by the way, could you explain to me how h = gt^2 is derived?
     
    Last edited: Dec 20, 2013
  14. Dec 19, 2013 #13
    I understand. But because I defined gravity to be negative if downwards,
    vi = Sqrt[-2(-)h]
    vi will give me a positive value.
     
  15. Dec 19, 2013 #14

    TSny

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    OK. That's fine. It's just not the usual way people use the symbol g.
     
  16. Dec 20, 2013 #15
    I know. I try not to make too much unnecessary arbitrary distinction in the equation. It's less chaotic this way.

    Can we continue?
     
  17. Dec 20, 2013 #16

    TSny

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    This is correct for the time to go from the ground up to maximum height.

    This is not a correct equation.
     
  18. Dec 20, 2013 #17
    Sorry. h = [gt^2]/2
    How is it derived?
     
  19. Dec 20, 2013 #18

    TSny

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    That was brought in by haruspex for his approach to the problem. I'll let him/her explain it. :smile:
     
  20. Dec 20, 2013 #19
    Capture.JPG
     
  21. Dec 20, 2013 #20

    TSny

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    Your result ##t = -\frac{\sqrt{-2gh}}{g}## is correct (assuming your convention that ##g## is a negative number).

    With the usual convention, you would have ##t = \frac{\sqrt{2gh}}{g}##. You can simplify the result by taking the ##g## in the denominator inside the square root and doing some canceling.
     
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