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?

Answer 1

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

Answer 2

I hope it was right