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Math Help - Finding Max height and time

  1. #1
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    Finding Max height and time

    The formula h = -16t^2 + v(subscript)0t + s(subscript)0 gives the height of an object tossed upward where v(subscript)0 represents the initial velocity, s(subscript)0 represents the initial height, and t represents time. A golf ball is hit straight up from the ground level with an initial velocity of 72 ft/sec. What is the maximum height that the ball reaches and the number of seconds it takes to reach that height?
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  2. #2
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    Quote Originally Posted by greenpumpkins View Post
    The formula h = -16t^2 + v(subscript)0t + s(subscript)0 gives the height of an object tossed upward where v(subscript)0 represents the initial velocity, s(subscript)0 represents the initial height, and t represents time. A golf ball is hit straight up from the ground level with an initial velocity of 72 ft/sec. What is the maximum height that the ball reaches and the number of seconds it takes to reach that height?
    You have: s_0 = 0 and v_0 = 72\ \frac{ft}{s}
    Plug in these values into the given equation:

    h(t) = -16t^2+72t

    This is the equation of a parabola opening down with it's maximum at it's vertex:

    h(t) = -16\left(t^2-\frac92 t+\frac{81}{16}\right)+81 = -16\left(t-\frac94\right)^2+81

    Therefore the coordinates of the vertex are V\left(\frac94\ ,\ 81\right)

    That means after 2.25 s the golf ball reaches it's maximum height of 81 ft.
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