Asked by Anonymous
If a rocket is propelled upward from ground level, its height in meters after t seconds is given by h=-9.8t^2+88.2t. During what interval of time will the rocket be higher than137.2m?
Do I set it up like this?
-9.8(1)^2+88.2(1)+137.2 = 215.6
Then h=b/2a=9.8/215.6=22m
137.2+22=159
Thanks for your help.
Do I set it up like this?
-9.8(1)^2+88.2(1)+137.2 = 215.6
Then h=b/2a=9.8/215.6=22m
137.2+22=159
Thanks for your help.
Answers
Answered by
Steve
Not quite. You found (sort of) when the height is greatest (the vertex of the parabola). That was not the question.
You have
h=-9.8t^2+88.2t
when is h=137.2?
(Aside. You sure about that 9.8? h = vt - 1/2 at^2 and a = 9.8)
137.2 = -9.8t^2 + 88.2t
t = 2,7
So, at those two times, h=137.2
Knowing what you do about the shape of parabolas, and that this one opens downward, h > 137.2 between those two values, so
2 < t < 7
You have
h=-9.8t^2+88.2t
when is h=137.2?
(Aside. You sure about that 9.8? h = vt - 1/2 at^2 and a = 9.8)
137.2 = -9.8t^2 + 88.2t
t = 2,7
So, at those two times, h=137.2
Knowing what you do about the shape of parabolas, and that this one opens downward, h > 137.2 between those two values, so
2 < t < 7
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