Algebra

The equation h= -16t^2 + 32t + 9 gives the height of a ball, h, in feet above the ground,at t seconds after the ball is thrown upward.How many seconds after the ball is thrown will it reach its maximum height? What is its maximum height?

asked by Anonymous
  1. v = dh/dt = 32 -32 t
    at top, v = 0
    t = 1
    h = 9+16 = 25
    ======================
    now with algebra
    16 t^2 -32 t =9 - h
    complete square
    t^2 -2 t = (-1/16)(h - 9)

    t^2-2t+1 = (-1/16)(h - 9)+16/16
    (t-1)^2 = -(1/16)( h-25)
    t = 1, h=25 same answer

    posted by Damon

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