Answer
a) $-5g$
b) $3g$
Work Step by Step
When the ball is in the air, there are two forces exerted on it which are its own weight due to Earth's gravitational pull and the drag force due to air resistance.
When it is moving up, these two forces are pointing downward.
And when it is moving down, the weight force is pointing down as usual but the drag force now is pointing up.
See the force diagrams below.
a) When the ball moves up:
In this case, the net force exerted on the ball is given by
$$\sum F_y=-mg-D=ma_y$$
$$-mg-D=ma_y$$
$$a_y=\dfrac{-\left[mg+\frac{1}{2}C\rho Av_i^2\right]}{m}\tag 1$$
since we know that the initial speed is twice the terminal speed.
The terminal speed is given by
$$\sum F_y=-mg-D=0$$
$$D=mg$$
$$\frac{1}{2}C\rho Av_T^2=mg$$
$$v_T^2=\dfrac{2mg}{ C\rho A}$$
$$v_T =\sqrt{\dfrac{2mg}{ C\rho A}}$$
and hence,
$$v_i=2v_T=2\cdot \sqrt{\dfrac{2mg}{ C\rho A}}$$
Therefore,
$$v_i^2= \dfrac{8mg}{ C\rho A}\tag 2$$
$$a_y=\dfrac{-\left[mg+\frac{1}{2}C\rho A \dfrac{8mg}{ C\rho A}\right]}{m}=\dfrac{-\left[mg+4mg \right]}{m}=\dfrac{-5mg}{m}$$
Therefore,
$$\boxed{a_y=-5g}$$
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b) When the ball moves down:
In this case, the net force exerted on the ball is given by
$$\sum F_y=D-mg=ma_y$$
$$D-mg=ma_y$$
$$a_y=\dfrac{\frac{1}{2}C\rho Av_i^2 - mg}{m} $$
since we know that the initial speed is twice the terminal speed. The terminal speed is given by equation (2) above.
$$a_y=\dfrac{\frac{1}{2}C\rho A\dfrac{8mg}{ C\rho A} - mg}{m} $$
$$a_y=\dfrac{ 4mg - mg}{m} =\dfrac{ 3mg }{m} $$
Therefore,
$$\boxed{a_y=3g}$$
