How long must escape velocity be maintained? Escape velocity from Earth's surface is 11.2 Kilometres/second.
How long would one need to maintain this escape velocity to actually escape Earth's gravitational pull? Must this 11.2 km/s velocity need to be maintained all the way up to geosynchronous orbit?
 A: No -  it's the energy you need to have to leave Earth completely, essentially how fast you need to fire a bullet for it to leave.
The idea of an escape velocity only comes about because current launch vehicles burn all their fuel very soon after launch and need to achieve a high speed to have enough kinetic energy to escape. If it was possible to build a rocket where you could have the engine running 'all the way up' you could reach space at any slower vertical speed you wanted. 
So if you had a ladder you could climb into space.  Although if you want to be in orbit at the top you do have to have enough tangential (sideways) velocity to maintain that orbit.
A: No, the escape velocity doesn't need to be maintained for any length of time. Escape velocity is the minimum speed you need to have at the Earth's surface to be able to escape the gravitational pull, without using a rocket or other continuous propulsion.
In other words, ignoring all sources of gravity other than the Earth, if you launch a projectile straight up at escape velocity from Earth's surface, it will slow down as it rises but will never quite stop. Any slower, and the projectile will stop at some point and fall back to Earth.
A: Related: Launch loop 
Assuming you are useing a rocket style launch vehicle, everyone has so far.
To reach orbit, the rocket must impart to the payload a delta-v of about 9.3–10 km/s. 
This figure is mainly (~7.8 km/s) for horizontal acceleration needed to reach orbital speed, but allows for atmospheric drag (approximately 300 m/s with the ballistic coefficient of a 20 m long dense fuelled vehicle), gravity losses (depending on burn time and details of the trajectory and launch vehicle), and gaining altitude.  
First stage - External Tank
Engines 3 SSMEs located on Orbiter
Thrust 5.45220 MN total, sea level liftoff (1,225,704 lbf)
Specific impulse 455 s
Burn time 480 s
Fuel LOX/LH2
Second stage - Orbiter
Engines 2 OME
Thrust 53.4 kN combined total vacuum thrust (12,000 lbf)
Specific impulse 316 s
Burn time 1,250 s
Fuel MMH / N 2 O 4.
Referances: Launch Vehicle , Orbital Spaceflight , Burn Time
A: Velocity equation of a body thrown straight up from Earth surface:
$v^2(h)=(v_o^2 – 2\bf{g}R)+2gR\frac{R}{h}$
where
$v_o$  - initial velocity, $R$ - Earth radius, $h$ - distance from  Earth center, $ \bf{g}$ - gravitational acceleration at $R$ 
If  initial velocity is escape velocity $v_o=v_e=\sqrt{2\bf{g}R}$ then
$v^2(h)= v_e^2\frac{R}{h}$
or
$v(h)= v_e\sqrt{\frac{R}{h}}$
A: Escape velocity is much easier to understand using the following definition.
For the example of earth:
Escape velocity is the velocity at which a body at rest at the edge of the universe would impact the earth given only the action of gravitational force between the earth and the body.
Since the action is reversible it stands that a body reaching escape velocity at the surface of the earth would come to rest at the edge of the universe.(Notwithstanding the expansion of the universe and other such matters)
A: The answer is not infinity, why, because the body was at rest at The Edge of earth !.
More.. having achieved velocity (speed) the object is in a state of acceleration why then at the moment of escape from gravity should the object not continue to accelerate upon its own acceleration as there in now nothing to hinder. I.E. continue through to the speed of light and possibly beyond. Interesting theory !
