Dropped in oil

With the numbers in the problem the flow around the ball bearings is laminar, and Stokes's law applies. The drag force on the ball bearing is $F=6\pi \eta r v= 3\pi \eta d v$ , where $d$ is the diameter of the ball bearing, $\eta$ is the viscosity and $v$ the velocity. For the dropped ball the drag force is vertical and the equation of motion is

$\ddot y= W-3\pi \eta d v=W-3\pi \eta d \dot y$

where $W$ is the weight of the ball bearing and the positive direction is taken downwards.

The drag force on the other ball is also opposite to the velocity, but the force is not vertical. The components of the force are $F_x=-F\frac{v_x}{v}$ and $F_y=-F\frac{v_y}{v}$ , where $v_x=\dot x$ and $v_y=\dot y$ are the horizontal and vertical components of the velocity, $v=\sqrt{v_x^2+v_y^2}$ and we selected the $x$ axis of the reference frame in the direction of the initial velocity. Since $F$ is proportional to $v$ , we can write the horizontal and vertical components of the force in a particularly simple way:

$F_x=-F\frac{v_x}{v}= -3 \pi \eta d \dot x$

$F_y=-F\frac{v_y}{v}= -3 \pi \eta d \dot y$

The equations of motion are

$\ddot x=-3\pi \eta d \dot y$

$\ddot y=W-3\pi \eta d \dot y$

Notice that the equation of motion in the vertical direction is exactly the same for the two balls. The the initial conditions are also the same (same vertical height, zero vertical velocity). Therefore they will move the same way and they reach the bottom at the same time.

NOTE: for turbulent flow, when the drag force is proportional to the square of the velocity, the separation of the horizontal and vertical motion is not possible.