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Kinematics kangaroo jump question

  1. Sep 18, 2005 #1
    a kangaroo jumps to a vertical height of 2.8m. how long was it in the air before returning to earth.

    I'm clueless on how to solve it. can someone help me please?

    thanks in advance

    a= -9.8 m/s^2 (since it's going the opposite direction of gravity, right?)

    i was wondering how you search for time with only these 2 quantities.
    Last edited: Sep 18, 2005
  2. jcsd
  3. Sep 18, 2005 #2
    All you need is the formula D=VT.

    Since you are trying to solve for T you just manipulate the formula to look like T= D/V.
    Then just plug in your numbers. Oh also, just incase you didn't know Velocity is the same thing as gravity. Hope this helps.

    EDIT:eek:ops... don't listen to my post.
    Last edited: Sep 18, 2005
  4. Sep 18, 2005 #3


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    No keita, that is not how you solve that problem. Velocity is not the same as gravity. "Gravity" is an acceleration, not a velocity.

    Intr3pid : Have you studied the equations of motion ? Any of them ring a bell ?
  5. Sep 19, 2005 #4
    thanks for suggestion, i already solved the problem!

    d=1/2at^2, since the initial velocity is 0!

    man i rattled my brain for a while trying to figure it out.
  6. Sep 19, 2005 #5


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    Intr3pid - If you solved it the way I think you did, think about it again. First off, if the initial velocity were zero, it would never leave the ground. Secondly, remember that for the entire trip, the displacement is 0.

    Now - the answer I think you have is related to the proper answer. Where during the kangaroo's trip will the velocity actually be 0? If you can answer that one, you should see an easy way to get the full answer to your question.
  7. Sep 19, 2005 #6

    sorry, i don't really get what you're saying. and my answer was wrong from the answers in the text.
  8. Sep 19, 2005 #7


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    Well, since it seems you've already checked it...

    The place where the kangaroo's velocity would be zero would be at the peak - the top of the arc. If you used g = 9.80 m/s^2 and s = 2.8m, what you found was the time for it to fall from the peak back to the ground. Symmetry arguments would tell you that this is exactly half of the time for the total trip - both up and down.

    Don't be discouraged by this. It takes a little time to start seeing problems like this in terms of the physical models used in kinematics. Mostly, you need to concentrate on what the concepts (such as displacement vs. distance, velocity vs. speed, and so on) actually mean and how those meanings are represented in the mathematical models you're building in class.
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