SOLUTION: A water rocket is launched upward with an initial velocity of 48 ft/sec. Its height h, in feet, after t seconds is given by h = 48t - 16t^2. After how many seconds will the height

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Question 283805: A water rocket is launched upward with an initial velocity of 48 ft/sec. Its height h, in feet, after t seconds is given by h = 48t - 16t^2. After how many seconds will the height of the rocket be 20 feet? Show all steps necessary to arrive at your solution.
Answer by dabanfield(803) About Me  (Show Source):
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A water rocket is launched upward with an initial velocity of 48 ft/sec. Its height h, in feet, after t seconds is given by h = 48t - 16t^2. After how many seconds will the height of the rocket be 20 feet? Show all steps necessary to arrive at your solution.
We need to solve:
20 = 48t - 16t^2
or
16t^2 - 48t + 20 = 0
Divide both sides by 4;
4t^2 - 12t + 5 = 0
(2t - 1)*(2t - 5) = 0
Solutions are:
2t - 1 = 0 and 2t - 5 = 0 or
t = 1/2 and t = 5/2
The first time is on the way up and the second is on the way down.