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An object is thrown vertically and has an upward velocity of 18 m/s when it reaches one fourth of its maximum height above its launch point. What is the initial (launch) speed of the object?

  • Physics -

    v = Vo - 9.8 t
    height versus time is a parabola
    h = Vo t - 4.9 t^2
    Vertex is at the top where v = 0 call that height capital H
    0 = Vo - 9.8 ttop
    ttop = Vo/9.8
    H = Vo ttop - 4.9 ttop^2
    H = Vo^2/9.8 - .5 Vo^2/9.8
    H = .5 (Vo^2/9.8)
    H/4 = .125 (Vo^2/9.8)
    at some time t it is at H/4 and v =18
    v = 18 = Vo - 9.8 t

    H/4 = Vo t - 4.9 t^2 = .125 Vo^2/9.8
    Vo = 18 + 9.8 t
    that gives two equations in the two unknowns, t and Vo
    solve for Vo

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