SOLUTION: A model rocket is launched with an initial velocity of 180 ft/s. The height, h, in feet, of the rocket t seconds after the launch is given by
h = −16t2 + 180t.
How many s
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Question 1003408: A model rocket is launched with an initial velocity of 180 ft/s. The height, h, in feet, of the rocket t seconds after the launch is given by
h = −16t2 + 180t.
How many seconds after launch will the rocket be 410 ft above the ground? Round to the nearest hundredth of a second.
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
h(t) = -16t^2 + 180t = 410
Solve for t
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