Seriously, maybe you intended to say that air resistance is not a factor, instead.
If we ignore drag caused by air resistance, and we're talking about planet Earth, then the acceleration due to the force of gravity relatively near the surface is constant: 32 feet/sec per second.
We could use the freefall formula, to answer the question in your exercise. The formula for freefall is often stated as a function of height in terms of elasped time:
\(\displaystyle h(t) = -\frac{1}{2} \cdot g \cdot t^2 + v_0 \cdot t + h_0\)
The symbol g is the acceleration due to gravity. It's 32 near the surface of Earth (when the distance units are feet and the time units are seconds).
The symbol v[sub:3il6ff56]0[/sub:3il6ff56] is the initial velocity, which is zero (in this exercise) because the object is not initially "thrown" downward causing it to move faster, but is simply released and allowed to begin the fall under nothing but its own weight.
The symbol h[sub:3il6ff56]0[/sub:3il6ff56] is the height of the object at time t = 0, so that's 11500 (in this exercise).
The symbol h(t) represents the height of the object after t seconds have elapsed.
Substituting the parameters 32, 0, and 11500 into the freefall formula gives a function for the height of your object at any number of elasped seconds, from the time it's released until the time it hits the ground.
\(\displaystyle h(t) = -16t^2 + 11500\)
The exercise asks for the difference between h(0) and h(20), yes?
