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Physics

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An object is thrown vertically and has an upward velocity of 5 m/s when it reaches three fourths of its maximum height above its launch point. What is the initial (launch) speed of the object?

  • Physics -

    total energy at beginning=totalenergy at point above
    1/2 m vi^2=1/2 m 5^2+mg(3/4 h)

    this indicates that you have two variables, height, and velocity initially.

    but we know in the final 1/4 h, vf=0, and

    Vf^2=Vi^2+2gd
    0=25-2*9.8*h/4
    or h=50/9.8
    now you can solve for vi in the first equation.

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