SOLUTION: h(t)= -16t^2+dt+c At time t=0, a rocket was launched upward from an initial height of 6 feet. Until the rocket hit the ground, its height, in feet, after t seconds was gi

Algebra ->  Algebra  -> Expressions-with-variables -> SOLUTION: h(t)= -16t^2+dt+c At time t=0, a rocket was launched upward from an initial height of 6 feet. Until the rocket hit the ground, its height, in feet, after t seconds was gi      Log On

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Question 630072: h(t)= -16t^2+dt+c

At time t=0, a rocket was launched upward from an initial height of 6 feet. Until the rocket hit the ground, its height, in feet, after t seconds was given by the function h aobve, in which c and d are positive constants. If the rocket reached its maximum height of 350 ft at time t=4, what was the height, in feet, of the rocket at time t=1?
(the answer is 140, by the way)
please include explanations!
Answer by stanbon(57262) About Me  (Show Source):
You can put this solution on YOUR website!
h(t)= -16t^2+dt+c
At time t=0, a rocket was launched upward from an initial height of 6 feet. Until the rocket hit the ground, its height, in feet, after t seconds was given by the function h aobve, in which c and d are positive constants. If the rocket reached its maximum height of 350 ft at time t=4, what was the height, in feet, of the rocket at time t=1?
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h(0) = -16*0^2 + d*0 + c = 6 feet
So, c = 6
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h(4) = 350
h(4) = -16*4^2+4d + 6 = 350
256 + 4d + 6 = 350
4d = 88
d = 22
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Equation:
h(t) = -16t^2 + 22t + 6
h(1) = -16 + 22 + 6
h(1) = 12 feet
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Cheers,
Stan H.