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Finding the height of a building

  1. Jan 19, 2009 #1
    1. The problem statement, all variables and given/known data

    Spiderman 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 interval of time of 1.2 s of his fall.

    What is the height h of the building?

    The acceleration is constant -g.

    2. Relevant equations

    y(t) = y(0) + v(0)t - 1/2gt^2
    v = v(0) - gt
    v^2 = v(0)^2 - 2g(y-y(0))

    3. The attempt at a solution

    I found the average velocity of the final interval of time to be (h/4)m / 1.2s = (h/4.8)m/s.

    I know the average velocity occurs at half the displacement; so, I used this velocity as the initial velocity for another displacement of h/8 (or half the distance of the final interval).

    Then I tried using g = -9.8m/s^2, v(0) = -(h/4.8)m/s, y(0) = (h/8)m, and y = 0 to find the time spiderman takes to travel that distance, but I cannot do that without figuring out h, my original target.

    Please help.
  2. jcsd
  3. Jan 19, 2009 #2


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    Write out equations for what you know.

    h = 1/2*g*T2
    T is your total time.

    0 = 1/4 h - V1/4 *(1.3) - 1/2*g*(1.2)2
    V1/4 is the velocity at 1/4 h.

    Since he started from rest:
    V1/4 = g*(T - 1.2)

    Solve for T and that will give you h won't it?
  4. Jan 20, 2009 #3
    I tried solving for T but I still got a wrong answer for h
  5. Jan 20, 2009 #4


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    0 = 1/4 h - V1/4 *(1.3) - 1/2*g*(1.2)2
    It should be
    0 = 1/4 h - V1/4 *(1.2) - 1/2*g*(1.2)2
  6. Jan 20, 2009 #5
    Saw that 1.3 too, but I changed it to 1.2 and it's still wrong. Maybe I'm not understanding the concept. How do I solve for T?

    Is it T = sqrt( 2h/g )?

    and v(1/4) = g * ( sqrt( 2h/g - 1.2 )?

    then solve for h. It's what I did, but I got 254m which is wrong.
  7. Jan 20, 2009 #6


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    Put h = 1/2*g*T^2 and V1/4 = g*( T-1.2) in the equation. You get
    1/4*1/2*g*T^2 - g*(T-1.2)(1.2) - 1/2*g*(1.2)^2 = 0
    Cancell g through out and solve for T.
  8. Jan 20, 2009 #7


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    Oops. Yes the 1.3 was a definite typo. Thanks for pointing that out rl.bhat.

    Solving the quadratic gives me a different number than 254.

    I can only suggest taking care with the signs and the algebra.
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