a) 
b) See interpretation below
Step-by-step explanation:
a)
The motion of both balls is a free-fall motion: it means that the ball is acted upon the force of gravity only.
Therefore, this means that the motion of the ball is a uniformly accelerated motion, with constant acceleration equal to the acceleration of gravity:

in the downward direction.
For the ball dropped from the initial height of
, the height at time t is given by
(1)
The ball which is thrown upward from the ground instead is fired with an initial vertical velocity
, and its starting height is zero, so its position at time t is given by
(2)
Therefore, the polynomial that represents the distance between the two balls is:

b)
Now we interpret this polynomial, which is:

which represents the distance between the two balls at time t.
The interpretation of the two terms is the following:
- The constant term,
, is the initial distance between the two balls, at time t=0 (in fact, the first ball is still at the top of the building, while the second ball is on the ground). For this problem, 
- The coefficient of the linear term,
, is the initial velocity of the second ball; this terms tells us that the distance between the two balls decreases every second by
feet.