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Falling object Kinematics motion

  1. Dec 12, 2013 #1
    1. The problem statement, all variables and given/known data

    A falling object travels one-fourth of its distance during the last second of its fall. From what height was it dropped?

    2. Relevant equations

    none

    3. The attempt at a solution


    yf - yi = vit + 0.5gt^2
    0.25x = 0 + 0.5(-9.8ms^-2)(1)^2
    x = 19.6m
     
  2. jcsd
  3. Dec 12, 2013 #2
    Are you sure that initial velocity is zero?
     
  4. Dec 12, 2013 #3
    It was not specified so I went in with the assumption vi = 0.
     
  5. Dec 12, 2013 #4
    The problem states that you have a falling object which has one more second to go (last second of it's fall). That means that from that point initial velocity isn't zero.
     
  6. Dec 12, 2013 #5
    Make sense-should I solve for vf first then?
     
  7. Dec 12, 2013 #6
    you have two unknowns (T & H), you need two equations.

    I would make them for two positions shown on a picture (H and 3/4H) by using above mentioned equation:

    H=Vo*t + 1/2*g*t^2
     

    Attached Files:

  8. Dec 12, 2013 #7
    H = vit + 0.5gt^2
    1/4H = vit + 0.5gt^2

    1/4H = vi(1) + 0.5g(1)^2
    vi = 1/4H + 4.9ms^-2
    H = (1/4H + 4.9ms^-2)t + 0.5gt^2
     
  9. Dec 12, 2013 #8
    Hello, please check attached photo. Does that make sense?
     

    Attached Files:

  10. Dec 12, 2013 #9
    Why is vi=0 along the 3/4H?
     
  11. Dec 12, 2013 #10
    Because object starts falling from height H (from rest) where he has initial velocity 0.
     
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