Answer
a) See the proof below.
b) $ v_x =v_0-\dfrac{ 6\pi \eta R }{m }x $
c) $1.06\;\rm m$
Work Step by Step
a)
We can use the chain rule to find the acceleration of the block (to prove the given formula).
We know that
$$a_x=\dfrac{dv_x}{dt}$$
Multiplying the right side by $dx/dx$
$$a_x=\dfrac{dv_x}{dt}\dfrac{dx}{dx}$$
Rearranging
$$a_x=\dfrac{dv_x}{dx}\overbrace{\dfrac{dx}{dt}}^{v_x}$$
$$a_x=v_x\;\dfrac{dv_x}{dx}\tag 1 $$
b) When the sphere is shot horizontally, there are only two forces exerted on it; the gravitational force downward and the drag force that has two components one is backward and the other is upward.
The direction of the drag force changes with time since it moves on a trajectory path at which its initial direction relative to horizontal is $\theta=0^\circ$
Thus, the net horizontal force is given by
$$\sum F_x=-D\cos\theta=ma_x$$
$$-D\cos\theta=ma_x$$
Recall that the drag force is given by $D=bv$ whereas $b=6\pi \eta R$.
$$-6\pi \eta R\;\;\overbrace{v\cos\theta}^{v_x}=ma_x$$
Noting that the direction of the velocity of the ball is the same direction of the drag force which means that both will make the same angle with the horizontal direction.
$$-6\pi \eta R v_x =ma_x$$
Plugging from (1);
$$-6\pi \eta R v_x =m v_x\;\dfrac{dv_x}{dx}$$
$$-6\pi \eta R =m \;\dfrac{dv_x}{dx}$$
Rearranging;
$$\dfrac{-6\pi \eta R }{m }dx=dv_x $$
Integrating;
$$\int_0^x\dfrac{-6\pi \eta R }{m }dx=\int_{v_0}^{v_x}dv_x $$
$$ \dfrac{-6\pi \eta R }{m }x= v_x-v_0 $$
$$ \boxed{v_x =v_0-\dfrac{ 6\pi \eta R }{m }x} $$
c) We need to solve the previous formula for $x$;
$$(v_x -v_0) \left[\dfrac{m }{- 6\pi \eta R }\right]=x$$
$$x= \left[\dfrac{(v_x -v_0)m }{- 6\pi \eta R }\right] $$
Plugging the given, and recall that the final velocity is zero since we need to find the distance when it stops.
$$x= \left[\dfrac{ -10\times10^{-2}\cdot 1.0\times10^{-3} }{- 6\pi \cdot 1.0\times10^{-3}\cdot 0.5\times10^{-2} }\right] $$
$$x= \color{red}{\bf 1.06}\;\rm m$$
