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What is the velocity at the midway point of a ball able to reach a height y when thrown with an initial velocity
(Assume the ball is thrown upward and that up is the positive direction. Use the following as necessary: y and g. )

Someone posted this question before and was given the answer

v= sqrt (v_o)^2 + 9.8y

This answer is apparently wrong.

  • Physics -

    Maybe it is a reading of the question. If the max height is y, midpoint is y/2

    If that is so, then the 9.8 should be -9.8, and the equation is dead on correct.

  • Physics -

    The motion to the highest point (height “H”)
    v = vₒ - g•t,
    v=0, t = vₒ/g.
    H = vₒ•t - g•t²/2 = vₒ• vₒ/g –(g/2) •(vₒ/g)² = vₒ²/2•g.
    Velocity at the midpoint (height “h“)
    v =vₒ - g•t1,
    t1 =(vₒ- v)/g.
    h = vₒ•t1 - g•t1²/2 =
    =vₒ•(vₒ- v)/g - g•(vₒ- v)²/2•g² =
    = [2•vₒ•(vₒ- v) -(vₒ- v)²]/ 2•g =
    ={2vₒ² - 2vₒ•v - vₒ² +2 vₒ•v - v²}/2•g =
    = (vₒ² - v²)/2•g.
    h=H/2 = vₒ²/2•2•g.
    (vₒ² - v²)/2•g = vₒ²/2•2•g.
    vₒ² = 2•v²,
    v =vₒ/√2 = vₒ/1.41 =0.707•vₒ

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