Question

The expression 16 represnts distance in feet an object falls after t seconds the object is dropped from a nnheight of 906 feet

Answer 1

You're close — the distance fallen after t seconds (in feet) is d(t) = 16t^2 (since d = 1/2·g·t^2 with g ≈ 32 ft/s^2).

If the object is dropped from a height of 906 ft, its height above the ground after t seconds is
h(t) = 906 − 16t^2.

It hits the ground when h(t) = 0:
906 − 16t^2 = 0 ⇒ t^2 = 906/16 ≈ 56.625 ⇒ t ≈ 7.53 seconds.

If you want, I can compute the velocity at impact or the distance fallen after a specific time.