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Math Help - Freefall / Gravity

  1. #1
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    Freefall / Gravity

    Hey guys! I have a question about one of these physics problems..

    The notorious 'Dr. B' is dropped from the top of a building by a group of disgruntled students. Some stories below, Dr. B's body is observed to pass by a window 2.0 m tall in 0.15 s. What is the average speed of Dr. B's body as it passes by the window?

    This is part one, and I'm sure it's 2.0/.15 = 13.33333333m/s, However, part two I'm really confused on what to do. It states:

    From how far above the window was Dr. B's body dropped?

    I have.. Final Velo = 13.33333333, time = .15s, I'm assuming initial velo = 0

    So I'm thinking I use
    d=(vi+vf)/2 *t where vi and vf are initial and final velocity.

    Doing that I get .9999999998. I'm assuming I need to use g = 9.80 somewhere in there.... Is that right or is there something I'm missing?

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  2. #2
    Senior Member vincisonfire's Avatar
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    You know  \Delta s = 2
    \Delta t = 0.15
    and of course  a = -g
    Now use \Delta s = v_i \Delta t + \frac{a \Delta t^2}{2} to find  v_i . Finally use  v_f^2 = v_i^2 + 2 a \Delta s to find the initial height. Note that  v_i becomes  v_f in the second formula.
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  3. #3
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    Hello, Kaln0s!

    Ah, the famous "falling body" problem . . .

    You're expected to know the "free fall" formula: . s(t) \:=\:h_o + v_ot - \tfrac{1}{2}gt^2

    where: . s(t) = height above level-zero at time t . . . and: \begin{Bmatrix} h_o &=& \text{initial height} \\ v_o &=& \text{initial velocity} \\ g &=& \text{acceleration} \end{Bmatrix}


    The notorious 'Dr. B' is dropped from the top of a building by a group of disgruntled students.
    Some stories below, Dr. B's body is observed to pass by a window 2.0 m tall in 0.15 s.

    (a) What is the average speed of Dr. B's body as it passes by the window?

    (b) From how far above the window was Dr. B's body dropped?

    Let h = height of the building.
    Since he was dropped, v_o = 0.
    The acceleraltion constant is 9.8 m/secē.
    . . Hence, the height function is: . s(t) \:=\:h - 4.9t^2 .[1]



    Let a = height of the top of the window.

    At time t, he is a feet above the ground.

    At time t + 0.15, he is a-2 feet above the ground.


    Substitute into [1]: . \begin{array}{cccc}h-4.9t^2 &=& a & {\color{blue}[2]} \\ h-4.9(t+0.15)^2 &=& a-2 & {\color{blue}[3]} \end{array}

    Subtract [3] from [2]: . 4.9(t+0.15)^2 - 4.9t^2 \:=\:2

    . . which has the solution: . t \:=\:\frac{1.88975}{1.47} \:\approx\:1.2855 seconds.

    Hence, it took 1.2855 second for the body to reach the top of the window.


    When t = 0, the height is: . h meters.

    When t = 1.2855, the height is: . h - 8.1 meters (approx.).


    Therefore, his body was dropped about 8.1 meters above the window.


    Maybe someone can find a shorter approach . . .
    .
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