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A Falling Stone

  1. Jun 2, 2005 #1
    Hi,
    I have a question about height and a falling stone. The question reads:

    Question: A stone falling takes 0.28 seconds to travel past a window 2.2 meters tall. From what height above the tope of the window did the stone fall?

    Answer: 3.46 meters

    Is this the correct answer.

    Thank You
     
  2. jcsd
  3. Jun 2, 2005 #2

    dextercioby

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    Homework Helper

    How did you do it?

    Daniel.
     
  4. Jun 2, 2005 #3

    DaveC426913

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    i.e. show your work if you want help.
     
  5. Jun 2, 2005 #4
    heh, i remember doing that problem out of giancoli years ago.
     
  6. Jun 3, 2005 #5
    Work

    Sorry here's my work.
    Work:

    t=0
    y(i)=0
    v=0
    V(i)=7.86 m/s
    a= -9.8 m/s^2



    y= (v^2-V(i)^2)/2a

    y=(0-(7.86)^2)/(2*.9.8)

    y=3.46 m

    Thank You
     
  7. Jun 3, 2005 #6
    What do you denote by v(i) and how did you derived it? If you mean the speed at the top of the window I assume you found it as:

    [tex]v_i = \frac{h - \frac{gt^2}{2}}{t}[/tex],

    Its value should be 6.49 m/s instead.

    Also do not consider accelearation as negative. It is easier to take the y-axis pointing downward to make all vector quantities positive.


    Almost there, but you need to arrange a little, as v_i is actually the final velocity for the part of the trajectory from drop point to the top of the window. So,

    [tex]h = \frac{v_i^2}{2g}[/tex]

    You should get a value of 2.15 m.
     
    Last edited: Jun 3, 2005
  8. Jun 3, 2005 #7
    Thank you, I see where I got confused
     
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