SOLUTION: A projectile is fired straight upward from ground level with an initial velocity of 80 feet per second. During which interval of time will the projectile's height exceed 96 feet?

Question 1074612: A projectile is fired straight upward from ground level with an initial velocity of 80 feet per second. During which interval of time will the projectile's height exceed 96 feet?
I got h=-16t^2+80t > 96
Can someone help me with this? Am I wrong here?
Thank you.
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
A projectile is fired straight upward from ground level with an initial velocity of 80 feet per second. During which interval of time will the projectile's height exceed 96 feet?
I got h=-16t^2+80t > 96
Can someone help me with this? Am I wrong here?
--------------
You're not wrong, but it's easier to solve the equation first.
h(t) = -16t^2+80t = 96
-16t^2 + 80t - 96 = 0
-t^2 + 5t - 6 = 0
t^2 - 5t + 6 = 0
(t - 2)*(t - 3) = 0
t = 2, 3
It's at 96 feet at t=2 second (ascending) and t=3 seconds (descending).
It's above 96 feet 2 < t < 3

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