SOLUTION: NASA launches a rocket at t=0 seconds. It's height, in meters above sea-level, as a function of time is given by h(t)= -4.9t^2+67t+339. How high above sea-level does the rocket get

Question 959793: NASA launches a rocket at t=0 seconds. It's height, in meters above sea-level, as a function of time is given by h(t)= -4.9t^2+67t+339. How high above sea-level does the rocket get at it's peak?
Answer by nerdybill(7384) (Show Source): You can put this solution on YOUR website!
h(t)= -4.9t^2+67t+339
The above is a "quadratic" (polynomial with a degree of 2).
If it is a quadratic, then the graph is in the shape of a parabola.
We know it opens downward because the coefficient associated with the x^2 term is negative (sad-face).
Since it opens downwards, the vertex is the MAX.
To find the 't' value of the vertex:
t = -b/(2a)
t = -67/(2*(-4))
t = -67/(-8))
t = 8.375 seconds.
.
to find the height, we plug into the original equation:
h(8.375)= -4.9(8.375)^2+67(8.375)+339
h(8.375) = 556.44 meters (answer)
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