classical mechanics - Again, why is kinetic energy and velocity independent of position coordinates in Cartesian coordinates

This might be a very simple question. I read one previous post Can the kinetic energy be a function of the position vector?

I know that in Cartesian coordinates, the kinetic energy $T=\frac{1}{2}mv^2$. And $T$ is not an explicit function of position. So $\frac{\partial T}{\partial x}=0$, where we suppose $x$ is a coordinate.

But I got confused by one example, we have a ball move vertically from the origin O, with a velocity of $\vec V_0$.

then when the ball reach $y_1$ in the positive $y$ axis, we have $$mgy=\frac{1}{2}mv_0^2-\frac{1}{2}mv_1^2$$ so the velocity at y is $$v_1=\sqrt {\frac{1}{2}mv_0^2-mgy}$$ Does it means that velocity and kinetic energy both are explicit function of position y in this case? I know this is a special case, but the statement that $\frac{\partial T}{\partial x}=0$ in Cartesian coordinates seems to be quite general. So where have I missed so far? Thanks guys!