Question 1009309
A rocket is shot into the air can be modeled by the equation y= -16x^2 + 420x + 250, where x represents the time the rocket is in the air and y represents the 
height of the rocket at the given time. 
What is the initial height of the object?:: f(0) = 250 ft.
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What is the initial velocity of the object?:: 420 ft/sec
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At what time does the rocket reach the ground?
Solve:: -16x^2 + 420x + 250 = 0
x = 26.83 seconds
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At what time does the rocket reach its maximum height?
Ans:: when x = -b/(2a) = -420/(-2*-16) = 13.125 sec
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What is the maximum height achieved by the rocket?
Ans: f(13.125) = 3006.3 ft
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At what time is the rocket at 200ft?
Solve -16x^2 + 420x + 250 = 200
x = 26.37 sec
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At what time is the rocket at 1500ft?
Solve -16x^2 + 420x + 250 = 1500
x = 3.42 sec and x = 22.83 sec
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What is the height of the rocket at 3.75 seconds?
f(3.75) = 1600 ft
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Cheers,
Stan H.
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