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Distance of Jump

  1. Mar 1, 2015 #1
    1. The problem statement, all variables and given/known data
    How high can a kangaroo jump. He is 66.5 kg, exerts 1881 N force on the ground when jumping, and reaches a velocity of 6.08 before leaving the ground (at 1 m). a=18.48

    2. Relevant equations
    F=ma

    3. The attempt at a solution
    looking for distance.
    vf^2=vi^2+2ax
    36.97=2(9.8)x
    x=1.88
     
  2. jcsd
  3. Mar 1, 2015 #2

    BvU

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    Oh boy, my story becomes repetitive. What's your question ? How can I help .... ?
     
  4. Mar 1, 2015 #3
    I am trying to find the highest distance Mr. Kangaroo can jump. But apparently my answer of 1.88 m is incorrect, and i do not understand where i went wrong.
     
  5. Mar 1, 2015 #4
    Pretty high....

    I got 1.88 seconds for the time to decelerate from 18.48 m/s², if that's what your x variable is... It's hard to tell...

    It sounds like at this point the kangaroo is 1 m high and travelling at 6.08 m/s...
    Initially accelerating at 18.48 m/s²

    What I would do though next time is add in the units for your values.
    It's hard to help people when they're not as specific as they can be.
     
    Last edited: Mar 2, 2015
  6. Mar 2, 2015 #5

    haruspex

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    The numbers you substituted in your equation are an acceleration of g, an initial speed of 0, and a final speed of 6.08m/s. What would a height calculated from those data mean? Have you used the information about the force and mass?
    It is not enough to remember an equation - you must also remember what relationship its variables must have in the real world for the equation to be valid.
     
  7. Mar 2, 2015 #6
    i thought the equation would give me his change in distance. i do not see where i can fit the force into the problem, which is why i am asking for help
     
  8. Mar 2, 2015 #7
    What do kangaroos do?

    14.jpg

    When things stretch, they have elas........
     
  9. Mar 2, 2015 #8

    BvU

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    Dear shoebox, again your equation only has F = ma. Not enough to calculate a distance, simply because there is no distance in the equation.
    (Distance is height in this exercise: we want to know how high the fellow jumps !)

    So what other equation is suitable to solve this problem ?
    Next step: collect the needed variable values (if there are variables that are not needed, that is not a disaster: exercise makers love to spice otherwise boring and obvious problems with noise and distractions :confused:) and calculate this distance/height.

    [edit] I see you already use another equation in your post #1. The 1.88 m is OK with me. So how high does he go (because he starts at ... m) ?

    Maybe its that simple, don't know.
     
  10. Mar 2, 2015 #9

    haruspex

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    Actually, forget my previous comments. I misinterpreted what you were trying to do.
    The information you are given about take off speed and distance travelled to that point make the mass and force information redundant. (Maybe this is just one part of a multipart question?)
    You only did one thing wrong - misunderstanding what was meant by height jumped in this question. It says the 6.05 m/s is achieved "at 1 m". In other words, the roo is considered to have jumped 1 m high at this point, even though it is still in contact with the ground. (I.e. it's looking at the vertical displacement of the mass centre from normal posture, not height of paws from the ground nor of head from the ground.)
     
  11. Mar 2, 2015 #10
    I sounds to me that the 1m is supposed to a be a spring extension... It says v=6.05 m/s at 1 m before the roo left the ground...
    Given that a roos legs are like springs, I think it would be appropriate that hooks law be used here.
    This would also make sense of the force on the ground.
     
  12. Mar 2, 2015 #11
    Actually word for word it was said...

    "reaches a velocity of 6.08 before leaving the ground (at 1 m). a=18.48"

    I could be wrong...
     
  13. Mar 2, 2015 #12

    haruspex

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    I don't see that your interpretation differs from mine.

    First, let's be clear how kangaroos hop (I write from hundreds of observations). They first crouch down a little, as you or I would. Having started hopping, they manage to store most of the downward KE as PE in their highly elastic leg tendons. At the lowest point, the mass centre is much the same height as when sitting because the tail is now off the ground. (But it's well below horizontal because more tendons in the rump store the tail's downward KE. The tip practically touches the ground.)

    Presumably, the question concerns rising from this lowest position. According to the question, a constant vertical force is exerted while contact with the ground is maintained (so, no, it's not a Hooke spring, though that would probably be closer to reality). If you use the given force and mass data, after the mass centre has risen 1m the velocity will be as given. From there it is only a matter of finding the additional distance to the highest point, which was done in the OP.
    The only tricky part is deciding whether "height of jump" is supposed to include the initial 1m. Seems like it is supposed to.

    Is that any different from your reading?
     
  14. Mar 2, 2015 #13
    No sir, you seem to have schooled me... I was imagining the 1m was the height he crouched from to spring up, rather the than the distance from his feet to his tail, still on the ground.

    I'm in Canada so not many kangaroos in these parts to observe :frown:

    ...-1881 is a ridiculous spring constant anyway....
     
  15. Mar 3, 2015 #14

    BvU

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    Dear shoebox,

    Can you post the exact wording of the exercise ? It looks as if the experts are discussing this over our heads :smile:

    Perhaps the answer is as simple as 1.88 + 1 ?
    Did you post because you didn't trust your answer or for another reason ?
     
  16. Mar 3, 2015 #15
    it is 1.88 plus one. I just forgot to add the 1m distance he had already covered
     
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