SOLUTION: The height in feet of an object thrown upward is given by the equation h = 12t + 9 – ½t2, where h is the height of the object after t seconds.
What is the initial position?
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Question 323983: The height in feet of an object thrown upward is given by the equation h = 12t + 9 – ½t2, where h is the height of the object after t seconds.
What is the initial position?
What is the height after ½ second?
At what time does it reach its maximum height?
What is the maximum height?
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
The height in feet of an object thrown upward is given by the equation h = 12t + 9 – ½t2, where h is the height of the object after t seconds.
What is the initial position?
What is the height after ½ second?
At what time does it reach its maximum height?
What is the maximum height?
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h = 12t + 9 – ½t2 A strange function, but could be true on a different planet.
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What is the initial position?
@ t=0: h(t) = 9 (no units were specified, 9 something)
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What is the height after ½ second?
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At what time does it reach its maximum height?
At the vertex of the parabola, t = -b/2a
t = -12/(-1) = 12 seconds
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What is the maximum height?
h(12) = (-1/2)*144 + 12*12 + 9
h(12) = 81
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