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?
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Anonymous

v = dh/dt = 32 32 t
at top, v = 0
t = 1
h = 9+16 = 25
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now with algebra
16 t^2 32 t =9  h
complete square
t^2 2 t = (1/16)(h  9)
t^22t+1 = (1/16)(h  9)+16/16
(t1)^2 = (1/16)( h25)
t = 1, h=25 same answerposted by Damon
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