A rocket is launched from a tower. The height of the rocket, y in feet, is related to the time after launch, x in seconds, by the given equation. Using this equation, find the maximum height reached by the rocket, to the nearest tenth of a foot.y, equals, minus, 16, x, squared, plus, 162, x, plus, 82y=−16x 2 +162x+82AnswerAttempt 1 out of 2
Question
A rocket is launched from a tower. The height of the rocket, y in feet, is related to the time after launch, x in seconds, by the given equation. Using this equation, find the maximum height reached by the rocket, to the nearest tenth of a foot.y, equals, minus, 16, x, squared, plus, 162, x, plus, 82y=−16x 2 +162x+82AnswerAttempt 1 out of 2
Solution
The equation given is a quadratic equation in the form of y = ax^2 + bx + c, where a = -16, b = 162, and c = 82.
The maximum height of the rocket is given by the vertex of the parabola represented by the equation. The x-coordinate of the vertex (h) can be found using the formula h = -b/(2a).
Substituting the given values, we get h = -162/(2*-16) = 5.0625 seconds.
Substitute h back into the original equation to find the maximum height (k):
k = -16*(5.0625)^2 + 162*5.0625 + 82 = 420.5625 feet.
So, the maximum height reached by the rocket is approximately 420.6 feet.
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