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How to find the time of two object passing each at opposite direction.

  1. Apr 29, 2012 #1
    1. A ball is dropped from a height of 79 m. Simultaneously a ball is thrown upward from the ground at a velocity of 28 m/s. How much time passes until the balls are at the same height.

    I need someone to check if my approach is correct.



    Position of falling object
    x =?
    Δt =?
    Xo = 79 m
    Vo = 0 m/s
    A = -9.8 m/s 2

    X = Xo + Vo Δt + .5a (Δt) 2
    X = 79 m + 0m/s (Δt) + .5(-9.8 m/s 2) (Δt) 2
    X = 79 m - 4.9 m/s 2(Δt) 2


    Position of elevating object
    x =?
    Δt =?
    Xo = 0 m
    Vo = 28 m/s
    A = -9.8 m/s 2

    X = Xo + Vo Δt + .5a (Δt) 2
    X (Δt) = 0m + 28m/s Δt + .5a (Δt) 2
    X (Δt) = 28 m/s (Δt) + .5(-9.8 m/s 2) (Δt) 2
    X (Δt) = 28 m/s (Δt) – 4.9 m/s 2 (Δt) 2

    To find the time, we need to set both the equation equal to each other.
    79 m – 4.9 m/s 2 (Δt) 2 = 28 m/s (Δt) - 4.9 m/s 2(Δt) 2
    79m = 28 m/s (Δt)
    2.8 s = Δt

    Then I replace 2.8 for t in both of the equation and check if my time is correct

    X = 79 m – 4.9 m/s 2 (Δt)2
    X = 79 m – 4.9 m/s 2 (2.8 s)2
    X = 79 m - 4.9 m/s 2 (7.84s 2)
    X = 79 m - 38.416 m
    X = 40.6 m

    X = 28 m/s (Δt) – 4.9 m/s 2 (Δt) 2
    X = 28 m/s (2.8s) - 4.9 m/s 2 (2.8 s)2
    X = 78.4 m - 38.416 m
    X = 40 m
    The answer is approximately the same.
     
  2. jcsd
  3. Apr 29, 2012 #2

    gneill

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    Staff: Mentor

    Your approach is fine, and the result of 2.8s is good (at least to one decimal).

    Keeping a few more decimal places in intermediate results would make your check of the answer more convincing :wink:
     
  4. Apr 29, 2012 #3
    So my answer is correct, right?
     
  5. Apr 29, 2012 #4

    gneill

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    Staff: Mentor

    I believe that is what I stated...
     
  6. Apr 29, 2012 #5
    Alrite thank you so much.
     
  7. Apr 29, 2012 #6

    gneill

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    Staff: Mentor

    No problem. Cheers :smile:
     
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