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?
--------------
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