The model for this type of vertical motion is h(t) = h(0) + v(0)t + (1/2)gt^2, where g is acceleration due to gravity, here -32 ft/(sec^2) because we're using the English system of measurement.
Inserting the given data results in:
5.5 ft = 2.5 ft + (45 ft/sec)t + (1/2)(-32 ft/(sec^2)t^2.
Note that this is a quadratic equation which can be solved in various ways. I'll use the quadratic formula.
Subtracting 5.5 ft from both sides, we get 0 = -3 + 45t - 16t^2. Then a = -16, b = 45 and c = -3. The discriminant, b^2 - 4(a)(c), is 45^2 - 4(-16)(-3), or 1833.
The solutions are as follows:
-45 plus or minus √1833 -45 plus or minus 42.8
t = --------------------------------------- = ------------------------------------
2(-16) -32
= 2.74 and a negative root.
Ignore the negative root, since we're measuring time, which is positive in this type of problem.
Thus, the ball will be in the air for 2.74 seconds.