Any launch profile will suffice (as long as it is not trying to go through the Earth) as long as the velocity at the end meet the following criteria, $$ \|\vec{v}\|>\sqrt{\frac{2GM}{\|\vec{r}\|}} $$ where $\vec{r}$ is the radius (position relative to the center of mass of the Earth) at that moment.

For this I also assume that its trajectory will not go through the Earth as well and is sufficiently out of its atmosphere.

But you could say it is about energy, since when escape velocity is reached the specific orbital energy becomes zero: $$ \epsilon = \frac{v^2}{2} - \frac{GM}{r} $$ because the gravitational potential is defined such that it goes to zero when $r$ approaches infinity. So at escape velocity if all kinetic energy would be converted into potential energy you would go infinitely far.

Any launch profile will suffice (as long as you do not try to go through the Earth of course) as long as the velocity at the end meets the following criteria, $$ \|\vec{v}\| \geq \sqrt{\frac{2GM}{\|\vec{r}\|}} $$ where $\vec{r}$ is the radius (position relative to the center of mass of the Earth) at that moment.

For this I also assume that its trajectory will not go through the Earth as well and is sufficiently out of its atmosphere.

You could few escape velocity from an energy point of view, since when escape velocity is reached the specific orbital energy becomes zero: $$ \epsilon = \frac{v^2}{2} - \frac{GM}{r}, $$ because the gravitational potential is defined such that it goes to zero when $r$ approaches infinity. So at escape velocity, if all kinetic energy would be converted into potential energy, then you would have to go infinitely far away.