SOLUTION: A model rocket is launched upward with an initial velocity of 270 feet per second. The height, in feet, of the rocket t seconds after the launch is given by h = −16t2 + 270t. H

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Question 1207285: A model rocket is launched upward with an initial velocity of 270 feet per second. The height, in feet, of the rocket t seconds after the launch is given by
h = −16t2 + 270t.
How many seconds after the launch will the rocket be 390 feet above the ground? Round to the nearest tenth of a second.

Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

h+=+-16t%5E2+%2B+270t
Now, we have to find out the time when h+=+390 feet.
390+=+-16t%5E2+%2B+270t
390+%2B16t%5E2+-+270t=0
16t%5E2+-+270t%2B390+=0
t=%28-%28-270%29%2B-sqrt%28%28-270%29%5E2-4%2A16%2A390%29%29%2F%282%2A16%29
t=%28270%2B-sqrt%2847940%29%29%2F32
t=%28270%2B-218.95%29%2F32

t=1.6 or t=15.3

answer is:
the rocket will be 390 feet above the ground
in t=1.6 seconds on the way up, and
in t=15.3 seconds on the way down