A rock is thrown straight up into the air from a height of 4 feet, the height of the rock above the ground, in feet, t seconds after it 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?

2.5 seconds.

Step-by-step explanation:

Find the value of t when the height is 28 feet:

-16t^2 + 56t + 4 = 28

-16t^2 + 56t - 24 = 0

-8 (2t^2 - 7t + 3) = 0

-8 (2t - 1) (t - 3) = 0

t = 0.5, 3 seconds.

So on the upward journey the rock is at 28 feet at 0.5 second after the throw and at 3 seconds it is at 28 feet again while it is falling back.

Therefore the period when it is at least 28 feet above the ground is 3.0 - 0.5 = 2.5 seconds.