a rock is thrown into the air from a height of 4 feet. the height of the rock above the ground, in feet, t seconds after the rock is thrown is given by -16t^2+56t+4. for how many seconds will the height of the rock be at least 28 feet above the ground

Answer 1

there will be a short time on the way up to 28 and a long one for the way down, we want the second minus the first

when will h = 28 feet?
28 = 4 + 56 t -16 t^2
16 t^2 - 56 t + 24 = 0
2 t^2 - 7 t + 3 = 0
(2t-1)(t-3) = 0
t = .5 or t = 3
so t = 2.5 seconds above 28