At a location where g = 9.8 m/s^2 , a solid sphere falls through...

When the sphere falls with a terminal velocity of $20 \text{ m/s}$, the net force acting on it is zero. Neglecting the small buoyant force in air, the upward drag force $F_d$ equals the downward gravitational force $mg$.

$F_d = mg = m \times 9.8$

In the gravity-free hall, $g = 0$, so the weight of the sphere is zero. When the sphere is pushed downward at $20 \text{ m/s}$, the same drag force $F_d$ acts on it in the upward direction, opposite to its velocity. The initial acceleration is given by:

$a = \dfrac{F_d}{m} = \dfrac{mg}{m} = g = 9.8 \text{ m/s}^2$ directed upward.

Since the drag force depends on the velocity of the sphere, as the velocity decreases, the drag force also decreases. Consequently, the magnitude of its acceleration will decrease over time.

The drag force continuously opposes the motion, reducing the velocity until the sphere eventually comes to a halt.