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Kinematics problems

  1. Aug 30, 2009 #1
    1. The problem statement, all variables and given/known data
    You throw a stone straight up with initial V of 24.5m/s and 2.0 s later you throw a second stone straight up with the same initial velocity. The first stone going down will meet the second stone going up. At what height above the point of release do the two stones meet?


    2. Relevant equations
    I'm not sure if I'm doing the question correctly. My attempt at it is below.



    3. The attempt at a solution
    1st Stone
    d = Vt
    d = 24.5t

    2nd Stone
    d = 24.5t + .5(9.8)t^2
    after 2 seconds, 24.5 x 2 = 49, so d + 49 = 24.5t +.5(9.8)t^2

    I then plugged in the d equation from the 1st stone
    24.5t + 49 = 24.5t + 4.9t^2
    4.9t^2 = 49
    t = 3.162 seconds is when the two stones are at the same distance.

    Then i assume that you just plug in the t-value in the distance equation for the 1st stone but i keep getting an incorrect answer. Any help please?
     
  2. jcsd
  3. Aug 30, 2009 #2

    tiny-tim

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    Welcome to PF!

    Hi marco101! Welcome to PF! :wink:
    Nope … that's for constant velocity, isnt it?

    You need an equation just like for the 2nd stone, but with an adjustment to t. :smile:
     
  4. Aug 30, 2009 #3
    Is my 2nd equation correct then?

    Im thinking if it's 2 seconds later, then the equation to the first stone is
    d = 24.5 (t+2) + .5(9.8)(t+2)^2

    while the 2nd stone stays the same, d + 49 = 24.5t +.5(9.8)t^2 and then set the two equal to each otehr?
     
  5. Aug 31, 2009 #4

    tiny-tim

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    Hi marco101! :smile:

    (just got up :zzz: …)
    Yup! :biggrin:
    Nooo … forget the 49 … they both start from d = 0 (one at t = 0, one at t = -2), don't they? :wink:
     
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