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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 out the time at which the rocket will reach its max, to the nearest 100th of a second. y=-16x^2+218x+75

User SpellTheif
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2 Answers

4 votes

Answer i got it wrong it was 5.94

Explanation:

User Ramesh Ponnusamy
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5 votes

Answer:

6.81s

Explanation:

First you must know that the velocity of the body is zero at the maximum height.

v =dy/dx

v= -32x+218

Since v = 0

-32x+218 =0

32x =218

x= 218/32

x = 6.81s

Hence the rocket reaches its maximum height after 6.81s

User TestTester
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