SOLUTION: A ball is thrown upward with an initial velocity of 32 ft/sec from a height of six feet. The function s(t)=-16t^2+32t+6 gives the height of the ball t seconds after release. Determ

Question 316048: A ball is thrown upward with an initial velocity of 32 ft/sec from a height of six feet. The function s(t)=-16t^2+32t+6 gives the height of the ball t seconds after release. Determine the time when the ball is at maximun height, and find that height.
Answer by scott8148(6628) (Show Source): You can put this solution on YOUR website!
s(t) is a downward opening parabola ("a" is negative) ; so the maximum is at the vertex , which is on the axis of symmetry

t = -b / 2a = -32 / 2(-16) = 1

s(1) = -16(1^2) + 32(1) + 6 = 22
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