#### Answer

(a) $a_y = \frac{5g}{4}$
(b) $a_y = \frac{3g}{4}$

#### Work Step by Step

The drag force $f$ is $Cv^2$, where C is a positive constant. The terminal speed $v_t$ is the speed when $Cv_t^2 = mg$.
$v_t = \sqrt{\frac{mg}{C}}$
(a) Suppose that $v$ is half the terminal speed.
$v = \frac{v_t}{2} = \sqrt{\frac{mg}{4C}}$
On the way up, the drag force $f$ is directed straight down. We can find the acceleration $a_y$.
$\sum F_y = m~a_y$
$mg+f = m~a_y$
$mg+Cv^2 = m~a_y$
$mg+C\frac{mg}{4C} = m~a_y$
$mg + \frac{mg}{4} = m~a_y$
$a_y = \frac{5g}{4}$
(b) Suppose that $v$ is half the terminal speed.
$v = \frac{v_t}{2} = \sqrt{\frac{mg}{4C}}$
On the way down, the drag force $f$ is directed straight up. We can find the acceleration $a_y$.
$\sum F_y = m~a_y$
$mg-f = m~a_y$
$mg-Cv^2 = m~a_y$
$mg-C\frac{mg}{4C} = m~a_y$
$mg - \frac{mg}{4} = m~a_y$
$a_y = \frac{3g}{4}$