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Distance and time

  1. Sep 5, 2006 #1

    888

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    hi,
    i nead help on solving a problem can someone help me please?
    if a kangaroo jumps a vertical height of 2.7m how long will it stay in the air before returning to the earth?

    thanks
     
  2. jcsd
  3. Sep 5, 2006 #2

    tony873004

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    What have you tried? What formula do you think you would use?
     
  4. Sep 5, 2006 #3

    888

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    i think i would use xf=xi+vit+.5at^2
     
  5. Sep 5, 2006 #4

    888

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    i tried using it but i got so confused
    there are other formulas i know but they dont have variable t for me to solve and there's vf=vi+at but i dont know vf
     
    Last edited: Sep 5, 2006
  6. Sep 5, 2006 #5

    tony873004

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    What will take longer? The time for the kangaroo to reach its maximum height, or the time for the kangaroo to drop from its maximum height back to Earth? What is the kangaroo's velocity when it is at its maximum height?
     
  7. Sep 5, 2006 #6

    888

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    it'll take longer to reach maximum height b/c of gravitational force and at max. height the vel is 0m/s
     
  8. Sep 5, 2006 #7

    888

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    so is the final vel. 0m/s? i thought the initial vel. was 0m/s?
     
  9. Sep 5, 2006 #8

    tony873004

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    Yes, at max height the velocity is 0. The initial velocity can not be 0 or the kangaroo would not leave the ground.

    It will not take longer to reach max height than it will to drop from max height back to the ground. They will be equal. This is very important because lots of the questions you will do depend on this.

    If time up = time down, then you only need to solve for one of them and then times it by 2 to get your answer. You're looking for time. The formula you gave me is correct. Can you use algebra to re-write that formula so it looks like " t = ...something " ?
     
  10. Sep 5, 2006 #9

    888

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    but im only finding the time that it last in the air so i dont need to multiply the answer by 2 rite?
     
  11. Sep 5, 2006 #10

    tony873004

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    It depends on what you compute. I find it easiest to compute how long it will take something to drop from rest from a height of 2.7 meters. But you could compute how long it would take something to ascend from the ground to a height of 2.7 meters as well. The kangaroo's total time in the air will be the time it took to rise to max height + the time it took to drop back to the ground from max height. So if you figure out one of these, and the other one is equal, you just multiply by 2.
     
  12. Sep 6, 2006 #11

    888

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    but dont i need to know what vi is or else i have two unknown variables in the equation t and vi
     
  13. Sep 6, 2006 #12

    tony873004

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    ** edit. Ignore this post. It was meant for another thread. Sorry...

    but if you write the equation for Earth, and set it equal to the equation for the Moon, things will cancel and you'll see this as a proportion rather than a formula that gives you an actual answer.

    That's all your final answer was. It was a proportion. It didn't matter what vi was for Earth. The same vi on the Moon got you 6 times higher.
     
    Last edited: Sep 6, 2006
  14. Sep 6, 2006 #13

    HallsofIvy

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    ?? What does the moon have to do with this??

    First, the initial velocity can't be 0 or the kangaroo will never get off the ground! The velocity of the kangaroo will be v= vi+ at at any time t (constant acceleration a so change in speed= at). The kangaroo will continue to go up until its velocity is 0 (of course, a= -9.8 m/s2 here). That is, the kangaroo will reach its greatest height when vi- 9.8t= 0 or t= vi/9.8. Yes, that depends on vi. Now put that for "t" in the height formula:xf= xi+ vit- 4.9t2. Taking xi= 0 (the kangaroo starts on the ground) and xf= 2.7 m, you have vi(vi/9.8)- 4.9(vi/9.8)2= 2.7. Solve that for vi and use vi to find the time t at which the kangaroo is at its highest point. As tony873004 told you, the kangaroo will take exactly as long to come down as to go up so its total time in the air is twice its time to the highest point.
     
  15. Sep 6, 2006 #14

    888

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    i'm having trouble rearranging the equation to get vi by itself and when i do i get the result of -518.62m/s as an answer for vi, can someone help me
     
  16. Sep 6, 2006 #15

    888

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    and when i plug it into the equation to get the time it turns out to be 100 and something which doesn't make sense
     
  17. Sep 6, 2006 #16

    tony873004

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    oops. I'm getting my posts mixed up. This was supposed to be in reply to the 'gravity help' thread. sorry.
     
  18. Sep 6, 2006 #17
    it's okay tony873004 we all make mistakes =)
     
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