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Kinmatics problem

  1. Sep 28, 2004 #1

    tony873004

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    A ball, dropped from rest, covers 2/7 of the distance to the ground in the last two seconds of its fall.

    (a) From what height was the ball dropped?
    (b) What was the total time of fall?

    If I can figure out either part a or part b, the other part will be easy.

    I have a feeling that the solution to this will involve simultaneous equations. But I don't really know where to begin aside from making a list of everything I know:

    [tex]d_{t} = d_{1} + d_{2}[/tex]
    [tex]d_{t} = \frac {2}{7} d + \frac {5}{7} d[/tex]

    [tex]t_{t} = t_{1} + t_{2} [/tex]
    [tex]t_{t} = t_{1} + 2 seconds[/tex]

    [tex]v_{i_{t}} = 0[/tex]
    [tex]v_{f_{t}} = v_{f_{2}}[/tex]

    [tex]d_{1} = \frac{5}{7}d_{t}[/tex]
    [tex]t_{1} = t_{t} - t_{2}[/tex]
    [tex]t_{1} = t_{t} - 2 seconds[/tex]

    [tex]v_{i_{1}} = 0[/tex]
    [tex]v_{f_{1}} = v_{i_{2}}[/tex]

    [tex]d_{2} = \frac{2}{7} d_{t}[/tex]
    [tex]t_{2} = 2 seconds[/tex]
    [tex]v_{i_{2}} = v_{f_{1}}[/tex]
    [tex]v_{f_{2}} = v_{f_{t}}[/tex]

    Can anyone suggest a starting point?
     
  2. jcsd
  3. Sep 28, 2004 #2

    Pyrrhus

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    Homework Helper

    Well, i will divide it in 2 parts, then i'll relate the equations with their link variable, the first part final speed, which will be the second part initial speed, Also i will use on the first part the equation without time, while on the second part i will use the displacement equation with time.
     
  4. Sep 29, 2004 #3
    Suppose at any time the y coordinate of the ball is y(t). Then the distance it has covered equals (h-y(t)). So,

    [tex](h-y(t = T)) - (h-y(t = T-2)) = \frac{2h}{7}[/tex]

    Additionally, y(t = T) = 0.

    Cheers
    vivek
     
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