Answer
See the detailed answer below.
Work Step by Step
Equations 8.4 are
$$a_x=-\dfrac{\rho CAv_x}{2m}\sqrt{v_x^2+v_y^2}\tag 1$$
$$a_y=-g-\dfrac{\rho CAv_y}{2m}\sqrt{v_x^2+v_y^2}\tag 2$$
which are the acceleration components of an object as a projectile subjected to drag force.
We know that the drag force is opposing the direction of the velocity of the object which means that it changes its direction as the direction of velocity changes.
The drag force is given by;
$$F_D=\frac{1}{2}\rho CAv^2\tag 3$$
where $\rho$ is the density of the air, $C$ is the coefficient of drag, $A$ is the cross-sectional area of the object, and $v$ is the object's velocity.
As we see in the figure below, at some point when the object is still moving up the curve of its projectile path at an angle of $\theta$, the forces exerted on it are its own weight due to Earth's gravitational pull and the drag force.
The net force exerted on the object in the $x$-direction is given by
$$\sum F_x=-F_D\cos\theta=ma_x$$
Thus,
$$a_x=\dfrac{-F_D\cos\theta }{m}$$
Plugging from (3);
$$a_x=\dfrac{-\rho CA v^2 \cos\theta }{2m}$$
where $\theta$ is the angle between the velocity vector and the $+x$-direction. So that, $\cos\theta=\dfrac{v_x}{v}$
Thus,
$$a_x=\dfrac{-\rho CA v^2 \dfrac{v_x}{v} }{2m}$$
$$a_x=\dfrac{-\rho CA v_x }{2m}v $$
where $v=\sqrt{v_x^2+v_y^2}$
$$\boxed{a_x=\dfrac{-\rho CA v_x }{2m} \sqrt{v_x^2+v_y^2}} $$
The net force exerted on the object in the $y$-direction is given by
$$\sum F_y=-F_D\sin\theta-mg=ma_y$$
Thus,
$$ a_y=\dfrac{-F_D\sin\theta-mg}{m}$$
Plugging from (3);
$$ a_y=\dfrac{-\frac{1}{2}\rho CAv^2\sin\theta-mg}{m}$$
$$ a_y=\dfrac{-\frac{1}{2}\rho CAv^2\sin\theta}{m}-\dfrac{\color{red}{\bf\not}mg}{\color{red}{\bf\not}m}$$
$$ a_y=-\dfrac{ \rho CAv^2\sin\theta}{2m}-g$$
where $\sin\theta=\dfrac{v_y}{v}$;
$$ a_y=-\dfrac{ \rho CAv^2\dfrac{v_y}{v}}{2m}-g$$
$$ a_y=-\dfrac{ \rho CAv_y}{2m}v -g$$
where $v=\sqrt{v_x^2+v_y^2}$
$$\boxed{ a_y=-\dfrac{ \rho CAv_y}{2m}\sqrt{v_x^2+v_y^2} -g}$$
