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Acceleration with limited known variables Question

  1. Sep 11, 2009 #1
    1. The problem statement, all variables and given/known data

    A bush baby, an African primate, is capable of leaping vertically to the remarkable height of 2.26 m. To jump this high, the bush baby accelerates over a distance of 0.160 m, while extending the legs. The acceleration during the jump is approximately constant. What is the acceleration in m/s?

    h=2.26m
    d=.160m
    g=9.8m/s

    2. Relevant equations

    d=.5 X a X t
    V=a X t
    h=V X t-1/2g(t^2)
    0=V - g X t


    3. The attempt at a solution

    I have no idea how to attempt this without another variable given.
     
  2. jcsd
  3. Sep 11, 2009 #2

    berkeman

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    Staff: Mentor

    Acceleration is in m/s^2, not m/s.

    The acceleration happens first, to give a Vo. Then the rest is just governed by the simple/regular gravity kinematic equations. Show your work.
     
  4. Sep 11, 2009 #3
    Here's what I got, and I submitted it and it's still wrong. Can anyone tell me where I am going wrong?

    The square of the max initial velocity Vo² = 2gh where h is 2.16 - .16 = 2.1 m.

    Vo² = 41.16 m²/s²

    To reach this speed in a jump of y meters, a = Vo²/(2y) = 41.16/(2*.16)

    a = 128.625 m/s = 13.125 g's
     
  5. Sep 11, 2009 #4

    berkeman

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    Staff: Mentor

    The numbers look right to me, except for the units on the final answer (I'll say again, the units of acceleration are m/s^2, not m/s). If you carried the units along in your equations above the answer, you would have gotten the same number and the correct units.

    If you submit a = 128.63 m/s^2, is it still wrong?
     
  6. Sep 11, 2009 #5
    Yeah, it's wrong. All I need is the numbers, the m/s^2 is given. It said the answer is 138! Anyone know how they got that?
     
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