A ball is shot from a compressed air gun at twice its terminal speed

The magnitude of the drag force is given by

Fd = -1/2*rho*v^2*Cd*A.

Where rho is the density of air, v is the velocity, Cd is the drag coefficient, and A is the cross-sectional area.

At the terminal velocity, vt, this will be equal and opposite to the gravitational force

1/2*rho*vt^2*Cd*A = m*g

Now, because the drag force is proportional to the square of the velocity, if we double the velocity the magnitude of the drag force will increase by a factor of 4.

Fd = 1/2*rho*(2*vt)^2*Cd*A
Fd = 1/2*rho*4*vt^2*Cd*A
Fd = 4*(1/2*rho*vt^2*Cd*A)
Fd = 4*m*g

So the drag force is 4 times the acceleration due to gravity, and

m*a = 4*m*g
a = 4*g

the acceleration is 4*g.

This drag force will always act in the direction opposite to the motion of the ball. When the ball is shot straight up gravity and the drag force will be acting in the same direction. So

a = 4*g + g
a = 5*g.

When the ball is fired downwards the drag force will be acting in the opposite direction to gravity.

a = 4*g - g
a = 3*g