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Homework Help: Free fall question

  1. Aug 6, 2004 #1
    I'm really bad at physics so please excuse me. But I would like some help on this question. I would like to be able to solve it without plugging in any numbers untill the end.

    If a person steps off a building of height h, and free falls on the way down to the bottom. What is the height of the building if he falls a distance of h/4 in his last 1 second of fall.

    Any ideas?
     
  2. jcsd
  3. Aug 6, 2004 #2

    Doc Al

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    What's the relationship between distance fallen and time for a falling object?
     
  4. Aug 6, 2004 #3
    Is it[tex]t=\frac {2 \Delta x}{ v+v_0}[/tex]?

    Then all I have to solve is
    [tex]\frac {h}{4} = \frac {1}{2}g \left(\frac {2 \Delta x}{ v+v_0} - 1 \right)^2 + v_0 \left(\frac {2 \Delta x}{ v+v_0}-1 \right)+h[/tex]

    Where [tex]v_0=0, \Delta x = \frac {3h}{4}[/tex]. Am I correct?
     
  5. Aug 6, 2004 #4

    Integral

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    Your basic problem is:

    [tex] x(t) = - \frac {g t^2} 2 + V_0 t + X_0 [/tex]

    [tex] V_0 =0 [/tex]
    [tex] X_0 =h [/tex]

    so you get an equation of motion as:
    [tex] x(t) = - \frac {g t^2} 2 + h [/tex]

    Solve for t when x=0, this will give you an expression, call it T, for the time of the fall in terms of g and h. Now you know that

    [tex] x(T-1) = \frac h 4 [/tex]

    use this in your equation of motion to find h.
     
  6. Aug 6, 2004 #5
    Oh I see, ok thanks a lot.
     
  7. Aug 7, 2004 #6
    Alternatively you can use the four basic constant acceleration equations in a simultaneous equation i think.
     
  8. Aug 8, 2004 #7
    You could also solve this way:

    let t = time it takes to fall.

    since x = .5gt^2 (one of the general motion equations)

    h = .5g(t^2)

    .75h = .5g(t - 1)^2 (since time will be t-1 when it has fallen .75h)

    just subtract one equation from another to get .25h = .5(t^2) - .5(t - 1)^2....and simplify to further get....

    ..... h/4 = 2g(2t - 1)....multiply by 4 to get:

    h = 2g(2t - 1)...and set this back equal to:

    .5g(t^2) = 2g(2t - 1).....this ends up becoming t^2 - 8t + 4 = 0....and then solve for t with the quadratic. Now that you know t, you can easily find h.
     
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