SOLUTION: A model rocket is launched with an initial velocity of 160 ft/s. The height, h, in feet, of the rocket t seconds after the launch is given by h = −16t2 + 160t. How many secon

Question 1139880: A model rocket is launched with an initial velocity of 160 ft/s. The height, h, in feet, of the rocket t seconds after the launch is given by
h = −16t2 + 160t.
How many seconds after launch will the rocket be 380 ft above the ground? Round to the nearest hundredth of a second.
Found 2 solutions by Alan3354, ikleyn:
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
when −16t2 + 160t = 0

Answer by ikleyn(52798) (Show Source): You can put this solution on YOUR website!
.

It will be at the given height when h(t) = 380, or, equivalently, = 380 = 0. Solve this quadratic equation = = . There are two solution, both make sense, = = 3.88 seconds in the increasing part of the parabola, and = = 6.12 seconds in the decreasing branch of the parabola.

Solved.

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