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A small rocket is launched from a height of 6 feet off of the ground at an initial upward velocity of 64 feet per second. Its height is given by the equation 2003-18-06-00-00_files/i0020000.jpg, where h is the height of the rocket in feet and t is the time since liftoff in seconds. At what time is the rocket 54 feet in the air?
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\[H(t) = -\frac{ 1 }{ 2 }(10)t^2 + v_{i}t + h_{i}\] The height a projectile is at is equal to -(acceleration of gravity) times time squared + the initial velocity multiplied by time plus the initial height.
Replace 10 with whatever # your teacher usually puts for gravity. Some people have different preference.
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