Question 430769: A projectile is launched upward from the ground. Its height s in feet above the ground after t seconds is given by
s= 320t-16t^2
a) After how many seconds in the air will it hit the ground? Round to the nearest hundredth.
b) During what time interval is the projectile more than 576 feet above the ground? Round to the nearest hundredth.
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! s= 320t-16t^2
a) After how many seconds in the air will it hit the ground? Round to the nearest hundredth.
It hits the ground at s = 0.
320t-16t^2 = 0
t = 0 (at launch)
t = 20 (at impact)
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b) During what time interval is the projectile more than 576 feet above the ground? Round to the nearest hundredth.
320t-16t^2 = 576
t^2 - 20t + 36 = 0
(t-2)*(t-18) = 0
t = 2,18
It's at 576 at t=2 going up, and at t=18 descending.
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