They are pretty much the same thing. If they are not, then then one is just a constant offset of the other.
When we talk about the position of an object, we usually express that position in terms of a position vector $\mathbf r(t)$. This vector can be expressed based on a coordinate system, for example in Cartesian coordinates
$$\mathbf r(t)=x(t)\hat x+y(t)\hat y+z(t)\hat z$$
Note that $\mathbf r(t)=0$ at the origin, i.e. when $x=y=z=0$. Additionally, you could draw the path this vector traces in time, and it would be seeing the position of the object at various times points. For your case $x(t)=z(t)=0$, and so you can easily plot $y(t)$ vs. $t$.
Displacement $\Delta\mathbf r(t)$ is the change in position of an object relative to some initial position $\mathbf r_0$. i.e.
$$\Delta\mathbf r(t)=\mathbf r(t)-\mathbf r_0$$
Note that displacement is $0$ when $\mathbf r(t)=\mathbf r_0$. Typically, we just choose $\mathbf r_0=0$ so that displacement and position are equal. However, you could instead use the displacement from some other position that is not the origin. Then displacement and position would not be the same thing, but they would just be offset by the constant position vector $\mathbf r_0$. Another way to look at this is that you are explicitly stating a shift in your coordinates so that the origin is at $\mathbf r_0$ compared to the original coordinate system.
With all of this in mind, physicists usually use position and displacement interchangeably, and they might not be as precise as what is discussed in this answer. However, what is actually meant is typically obvious from the context. I would say it is a distinction that you should not really worry about.