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Balloon falling

  1. Feb 15, 2010 #1
    As you look out of your apartment window, you see water balloons falling past. You measure that the balloons are visible (going past the window)for a time t, and the vertical length of your window is L_w. assume that the water balloons were dropped (rather than thrown downward). Determine then height h (above the top edge of your window) from which it was dropped. This will be an algebraic expression in terms of t, L_w, and g.

    my approach:
    known's:
    a_y= -g
    Vi_y= 0
    Vf_h=Vi _(L_w)

    Picture I made to help with solution
    http://yfrog.com/5hbaloonfallingj

    My use of kinematics
    Vf _h= -g*t which also means Vi_(L_w) = -g*t
    Yf_h = Yi_(L_w)
    L_w = Yi_(L_w) - g*t*t - (1/2)g*t^2 ------> Yi_(L_w) = L_w +(1/2)g*t^2
    Yf_h = Yi_h - (1/2)g*t^2 ----> L_w +(1/2)g*t^2 = Yi_h -(1/2)g*t^2 --(solve for Yi_h)--> Yi_h = g*t^2 + L_w
    Answer: [Yi_h = g*t^2 + L_w] is the algebraic expression to find height above the top of window edge.

    This is what I came up with and wondering if this is the correct answer for this problem?
     
  2. jcsd
  3. Feb 16, 2010 #2

    ehild

    User Avatar
    Homework Helper
    Gold Member

    No, it is not correct. Try to write your derivation a bit more clearly. First, choose your reference system with the y axis pointing downward and the origin at the point from where the balloon was dropped. The y coordinate of the top of the window is y1=h. The bottom is at y2=h+L.

    At time t1 the balloon is at

    [tex]h=\frac{1}{2}t_1^2[/tex]

    and at time t2 it is at


    [tex]h+L=\frac{1}{2}t_2^2[/tex]


    ehild

    You measure the time span while the balloon moves from h to h+L, this is t2-t1=t. So t2=t1+t.

    Insert this for t2 in the second equation, determine t1, plug into the first equation: you get h.
     
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