# math

A model rocket is launched with an initial velocity of 250 ft/s. The height, h, in feet, of the rocket t seconds after the launch is given by
h = −16t2 + 250t.
How many seconds after launch will the rocket be 730 ft above the ground? Round to the nearest hundredth of a second.

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1. solve
730 = -16t^2 + 250t

16t^2 - 250t + 730 = 0

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posted by Reiny

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