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How fast would a ball have to be thrown upward to reach a maximum height of 49 ft? [Hint: Use the discriminant of the equation
16t2 − v0t + h = 0.

  • math -

    This equation is derived from the following:

    at maximum height the velocity is zero.
    y(current height)= y0 (initial height)+v0t+.5at^2

    take the derivative

    32t-v0=0 so vo=32t

    Plug this in for the v0 value

    16t2 − (32t)t + h = 0

    h=49ft

    -16t^2+49=0

    Solve for t:
    16t^2=49
    4t=7
    t=7/4

    Plug in t and h to solve for v0

    16t2 − v0t + h = 0 where h=49 and t=7/4

    16(7/4)^2-v0 (7/4) +49=0

    :P

  • math -

    How fast would a ball have to be thrown upward to reach a maximum height of 49 ft? [Hint: Use the discriminant of the equation 16t2 − v0t + h = 0.]

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