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Calculating acceleration due to gravity on an alien planet

  1. Jun 1, 2009 #1
    1. The problem statement, all variables and given/known data
    On an alien planet you throw a stone upward and measure the times Ta and Tb. The difference in heights of point a and point b is h. Calculate the acceleration due to gravity.
    Ta = 7.1s; Tb= 4.4s; h = 15.1m


    2. Relevant equations
    a = dv/st
    v(t) = vo - gt


    3. The attempt at a solution
    I have tried solving a = t*(2gh)^.5. I have also tried a = 2h/(Ta-Tb). I keep getting numbers that are far too great I believe based off of the numbers I am given.
     
  2. jcsd
  3. Jun 2, 2009 #2

    Hootenanny

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    I'm afraid that you cannot use either of those two equations, since you do not know that initial and final velocities of the stone. And what is that g doing in your attempt at the solution? Here g and a are the same thing.

    Can you think of a better kinematic (SUVAT) equation to use rather than the one that you have chosen?
     
  4. Jun 2, 2009 #3
    I don't understand how you are suppose to use kinematic equations when you do not have any of the variable to work with.
     
  5. Jun 2, 2009 #4

    Hootenanny

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    Well that not quite right, we do no some things. Let's make two lists. First, write down that variable that you want to find out. Second, write down a list of everything that you know.
     
  6. Jun 2, 2009 #5
    okay so we have two time variables, a height displacement and we want to find acceleration. But most equations ask for velocity, which I do not know. So do I need to solve for the velocity at each time given. Then find the difference between those velocities and then divided that by the difference in time?
     
  7. Jun 2, 2009 #6

    Hootenanny

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    Good. So now list the (four) kinematic equations for constant acceleration.
     
  8. Jun 2, 2009 #7
    v=at+vo
    dv=.5(v+vo)
    x=.5at^2 + vot + xo
    x=dvt + xo = (v^2 - vo^2)/2a + xo
     
  9. Jun 2, 2009 #8

    Hootenanny

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    Good. So, we know x, t and v1. We want to find out a, but we don't know v0.

    Can you use two of the above equations to eliminate v0?
     
  10. Jun 2, 2009 #9
    I get: .5at^2 + xo = x
     
  11. Jun 2, 2009 #10

    Hootenanny

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    Good.

    Can you now solve the problem?
     
  12. Jun 2, 2009 #11
    the only question i have left is: x is equal to h and i assume xo=0 right?
     
  13. Jun 2, 2009 #12
    another question: for the time is it the time difference?
     
  14. Jun 2, 2009 #13

    Hootenanny

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    Correct.
    Yup :smile:
     
  15. Jun 2, 2009 #14
    okay so with numbers i get: (2*15.1)/((7.1-4.4)^2) = 4.143 which is not correct. what am i doing wrong?
     
  16. Jun 2, 2009 #15
    okay I think I am doing something wrong since I keep getting vt=0.
     
  17. Jun 2, 2009 #16

    Hootenanny

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    I don't see any reason why your answer should be wrong, except perhaps that it may need a negative sign since the acceleration is vertically downwards.
     
  18. Jun 2, 2009 #17
    i have tried -4.143 m/s^2 and it is still wrong
     
  19. Jun 2, 2009 #18

    Hootenanny

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    Does the online submission program specify a required accuracy?
     
  20. Jun 2, 2009 #19
    i did finally get the answer. Thank you for all your help.
     
  21. Jun 2, 2009 #20

    Hootenanny

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    My pleasure :smile:
     
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