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Kinematics - What do these variables mean?

  1. Sep 20, 2007 #1


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    1. The problem statement, all variables and given/known data
    first off, i dont completely understand what do the variables in the equation do:


    the exercise:
    A body is thrown up vertically, it passes trough the height 9,8m twice. Time between those two passes is 4 seconds. Whats the starting velocity ?

    2. Relevant equations

    3. The attempt at a solution

    i presume i have to do it with two equation systems? first off i'd have to calculate the starting velocity needed to get up and down in four seconds. Then i'd calculate the starting speed needed to get to the "seconds starting speed" at 9,8m and that would be the answer.

  2. jcsd
  3. Sep 20, 2007 #2


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    Homework Helper

    [tex]x=x_0 + v_0 t + \frac{1}{2} a t^2[/tex]
    [tex]x[/tex] is displacement at time [tex]t[/tex]
    [tex]x_0[/tex] is initial displacement
    [tex]v_0[/tex] is initial velocity
    [tex]a[/tex] is acceleration (constant)
    [tex]t[/tex] is time

    hint: [tex]a=9.8 m/s^2[/tex] in this case
  4. Sep 20, 2007 #3


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    Homework Helper

    oh.. new to PF.. welcome :smile:
  5. Sep 20, 2007 #4


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    great thanks for explaining the equation for me (y) everything seems much clearer now.

    So i tried to solve it, but the answer isnt what its supposed to be =/ I mean, it different in the end of the book. Is the book wrong or am i ?

    9,8=0 + Vo4 - 1/2*9,8*4[tex]^{2}[/tex]

    4Vo = 1/2*9,8*4[tex]^{2}[/tex] + 9,8

    Vo= (1/2*9,8*4[tex]^{2}[/tex] + 9,8) / 4 = 22,05 m/s

    According to the textbook the answer should be 24 m/s

    Thanks again
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