Answer
a) $v_x =v_0e^{\frac{-6\pi\eta R t}{m}} $
b) $60.7\;\rm s$
Work Step by Step
a) 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
$$\sum F_x=-D\cos\theta=ma_x$$
Thus,
$$a_x=\dfrac{-D\cos\theta}{m}$$
Recall that the drag force is given by $D=bv$.
$$a_x=\dfrac{-bv\cos\theta}{m}$$
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.
Thus, $v_x=v \cos\theta$ and at $t=0$, $\theta=0$ and hence, $v_x=v $.
So,
$$a_x=\dfrac{-b\;\;\; \overbrace{v \cos \theta}^{v_x}}{m}=\dfrac{-bv_x}{m}$$
Recall that $a_x=dv_x/dt$
$$\dfrac{dv_x}{dt}=\dfrac{-bv_x }{m}$$
Rearranging;
$$\dfrac{dv_x}{v_x}=\dfrac{-b }{m}dt$$
Integrating;
$$\int_{v_0}^{v_x}\dfrac{dv_x}{v_x}=\int_0^t\dfrac{-b }{m}dt$$
$$\int_{v_0}^{v_x}\dfrac{dv_x}{v_x}=\dfrac{-b }{m}\int_0^tdt=\dfrac{-b }{m}t$$
$$\ln v_x\bigg|_{v_0}^{v_x}=\dfrac{-b }{m}t$$
$$\ln v_x-\ln v_0=\ln\dfrac{v_x}{v_0} =\dfrac{-b }{m}t$$
$$\ln\dfrac{v_x}{v_0} =\dfrac{-b }{m}t$$
$$\dfrac{v_x}{v_0} =e^{\frac{-b t}{m}}$$
$$ v_x =v_0e^{\frac{-b t}{m}} $$
Plugging $b$ from the given;
$$\boxed{ v_x =v_0e^{\frac{-6\pi\eta R t}{m}}}$$
---
b) We need to find the time the ball takes to reach $v_x=0.5v_0$; Thus,
$$ \frac{1}{2}v_0 =v_0e^{\frac{-6\pi\eta R t}{m}} $$
$$ \frac{1}{2} = e^{\frac{-6\pi\eta R t}{m}} $$
$$\ln\left[ \frac{1}{2} \right]=\ln 1-\ln 2= {\dfrac{-6\pi\eta R t}{m}} $$
$$-\ln 2= \dfrac{-6\pi\eta R t}{m} $$
$$ \ln 2= \dfrac{ 6\pi\eta R t}{m} $$
Solving for $t$;
$$ t= \dfrac{m\ln 2} { 6\pi\eta R} $$
Plugging the given;
$$ t= \dfrac{33 \times10^{-3}\ln 2} { 6\pi\cdot 1.0\times10^{-3}\cdot 0.02 } =\color{red}{\bf 60.7}\;\rm s$$
