SOLUTION: A model rocket is launched off the top of a platform. The height of the rocket is given by the function h(t)= -4.9t^2 + 86t + 2.1, where h(t) is the height in meters and t is the t
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-> SOLUTION: A model rocket is launched off the top of a platform. The height of the rocket is given by the function h(t)= -4.9t^2 + 86t + 2.1, where h(t) is the height in meters and t is the t
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Question 322602: A model rocket is launched off the top of a platform. The height of the rocket is given by the function h(t)= -4.9t^2 + 86t + 2.1, where h(t) is the height in meters and t is the time in seconds.
A) When will the rocket's height be 200 m?
B) How long will it take for the rocket to reach the ground? Answer by galactus(183) (Show Source):
You can put this solution on YOUR website! For part A, just set the given quadratic equal to 200 and solve for t.
In other words, solve
You will get two solutions. One for when the rocket is going up and reaches 200 m, and the other when it is on its way down and descends to 200 m. I assume they mean the former (when it is ascending).
For part B, set the given quadratic equal to 0 and solve for t.
As an added bonus, find the maximum height of the rocket and how long it takes to get there.
You can differentiate the given quadratic or use t=-b/(2a)
-86/(2(-4.9))=8.775 seconds
The max height is then found by plugging back into the quadratic:
-4.9(8.775)^2+86(8.775)+2.1=379.45 m
Note that if we double 8.775, we get the same solution as in part B because the max height is halfway to the point where it lands. Very close, due to rounding.
