Relationship between depth and the vertical displacement of the object between two frames
Yes.  This all comes down to some simple trigonometry.  An object that is at horizontal distance $x$ and vertical distance $y$ from the camera's viewing direction will be at angle $\tan^{-1} (y/x)$ from the viewing direction, and the distance to the object will be the length of the hypotenuse, i.e., $\sqrt{x^2+y^2}$ (draw a picture).

So, if the object starts at height $y_1$ and then moves upward to height $y_2$, the distance to the object will increase from $\sqrt{x^2+y_1^2}$ to  $\sqrt{x^2+y_2^2}$.  Now you can compute the absolute or relative change in the distance to the object, if you know $x,y_1,y_2$.  For instance, if you know the distance to the object originally and the original angle $\theta_1$ from the camera's viewing direction, you can compute $x,y_1$; if you know how much it moved upward, you can compute $y_2$; and that will let you compute the new distance.