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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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?
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
