How can escape velocity be independent of the direction of projection of a body? I read that the escape velocity of a body is independent of the direction of projection.
For example, I could throw a ball at $11.2$ km/s velocity horizontally, and it would still leave earth.
I am unable to visualise or understand this. Could someone explain?
Edit: I think I should put my exact problem down more clearly.

At point A: I release the ball. It has horizontal velocity of 11.2 km/s.
From A to B: Gravity acts on the ball. One component slows down the horizontal velocity and one component constantly attracts the ball to earth.
At B: the ball is trying to leave earth with less than escape velocity.
What I realised: B might not be exactly on earth, the ball might have already covered some height.
So right now, I would just like someone to confirm if what I realised is right. (Or correct me.)
Thank you.
 A: Because it is a energy balance between a potential $V(r)$ and the kinetic energy of the object $ T = 1/2mv^2$, so the work to do provide in order to escape: $T > V$ does not depend on the path followed by the object (it only depends on the module of it’s velocity).
A: Ignoring atmospheric friction and buoyancy, if you threw the ball at a and it arrived at b, the velocity you threw the ball at would have to be about 8000 m\s, and the ball's velocity at b would not have changed.
If you threw the ball at escape velocity it would follow a parabolic path, with the vertex of the parabola at the throwing position.  The distance of the ball from the center of the Earth would be constantly increasing, and the velocity would be constantly decreasing, without ever reaching zero.
A: Of course, the escape velocity is all about energy, and a very special 'thing' about $1/r^2$ forces is that energy of an orbit does not depend on the eccentricity. Hence, if you are at a $r_0$ with energy $V(r_0)<0$, you just need:
$$ T(\vec v) = T(|v|) =\frac 1 2 mv^2 \ge -V(r_0) $$
to escape.
In principle, the independence of $E$ from $\epsilon$ can be related to the conservation of the Laplace-Runge-Lenz vector:
$${\bf A} = {\bf p}\times {\bf L} -mk{\bf {\hat  r}}$$
but at this moment, I am not having a "Eureka" moment as to how that applies here.
