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Building Height (HELP NEEDED)

  1. May 10, 2009 #1
    1. The problem statement, all variables and given/known data
    Spider-Man steps from the top of a tall building. He falls freely from rest to the ground a distance of h. He falls a distance of h/4 in the last 1 second of his fall. What is h?


    2. Relevant equations
    All kinematic equations with free fall.


    3. The attempt at a solution
    I separated the motion into 2 cases where there was one part three times higher than the other. I made a bunch of equations that lead to nothing. Apparently I don't understand this.
     
  2. jcsd
  3. May 10, 2009 #2

    diazona

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    Which equations did you make? How do they lead to nothing?
     
  4. May 10, 2009 #3
    Is it 276m?

    What I did:

    Let l = h/4

    The Kinematics equation s = vit + 1/2 *at2

    -l=v3l(1s) - 5*(1)2

    gives (5-l) = v3

    The distance spiderman moves before this is 3l, so using

    vf2=vi2+2as

    Vi is 0 since he starts from rest

    so (5-l)^2=2*3*g*l

    gives the quadratic equation l2-70x+25=0

    Solving gives x = 0.358984 or 69.641

    Multiply by 4 to get h.

    I'm not sure if it's right though :uhh:
     
  5. May 10, 2009 #4

    LowlyPion

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    I think you just need the right 2 equations.

    Focus on the time to fall. It falls 3/4 of the way in T1.

    3/4*h = 1/2*g*T12

    and

    h = 1/2*g*T22

    where T2 = T1 + 1

    Solve for the total time and it's all down hill from there.
     
  6. May 10, 2009 #5
    Thanks guys! That was a nice way to this problem LowlyPion. Also thanks physicsnoob93. I got 273m as my answer. I am pretty sure that's right after checking. Again thanks!
     
  7. May 10, 2009 #6

    diazona

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    Not that it matters now, but I tried it both ways and they both give the same answer, 273m...
     
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