SOLUTION: A stone is thrown directly upward from a height of 30ft with a velocity of 60ft/sec. The height of the stone t seconds after it has been thrown is given by the function s(t)= - 16t

Algebra ->  Rate-of-work-word-problems -> SOLUTION: A stone is thrown directly upward from a height of 30ft with a velocity of 60ft/sec. The height of the stone t seconds after it has been thrown is given by the function s(t)= - 16t      Log On


   

Question 451656: A stone is thrown directly upward from a height of 30ft with a velocity of 60ft/sec. The height of the stone t seconds after it has been thrown is given by the function s(t)= - 16t^2 + 16t + 30. Find the time at which the stone reaches its maximum height and find that maximum height.
Answer by nerdybill(7384) About Me  (Show Source):
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A stone is thrown directly upward from a height of 30ft with a velocity of 60ft/sec. The height of the stone t seconds after it has been thrown is given by the function s(t)= - 16t^2 + 16t + 30. Find the time at which the stone reaches its maximum height and find that maximum height.
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Time reaching maximum height is at the vertex:
t = -b/(2a)
t = -16/(2*(-16))
t = -16/(-32)
t = 1/2 seconds
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Maximum height:
s(1/2)= - 16(1/2)^2 + 16(1/2) + 30
s(1/2)= - 16(1/4) + 8 + 30
s(1/2)= - 4 + 8 + 30
s(1/2)= 34 feet