Rocket launch from a mountain If we were to build a high speed rail up the side of a mountain like in some Science Fiction movies, what is the velocity needed at the point of leaving the mountain excluding angular momentum from Earth’s rotation to achieve escape velocity?
 A: Escape velocity has got nothing to do with mass of the rocket. 
$v_e = \sqrt{\frac{2GM}{(r_0+h)}}$
$r_0 -$ Radius of the earth
$h -$ Height of the mountain
A: The total escape velocity is about 11,200 m/s (approximately 7 miles/sec.) in a direction roughly tangential to the earth's surface.  At 30 degrees north latitude (e.g., somewhere in southern Texas) the eath's spin would contribute about 400 m/s to the tangential velocity, so actual speed, relative to the earth's surface at said launch point would be about 10,800 m/s, if the load was launched in a roughly tangential direction, eastward; closer to the 11,200 m/s if you went straight up with it. 
The height of the mountain will serve a good purpose in getting you up to where the air is thinner when you leave the track; but, otherwise has negligible on how fast you need to go.
The main problem with this is that 11,200 m/s, at 14000 feet, is about Mach 24; roughly 13 times the muzzle velocity of a 30-06 high-powered rifle.  Heating, shredding, melting, ablation, and all sorts of bad stuff occur at these speeds, even in thin air.  
Studies have been done on this, to engineer a ground-power-assisted launch.  The idea is to get the load up to a very high speed --but not high enough to shred or melt it-- and let the rocket engines take over from there.  I believe I read about a track hypothesized for somewhere in the vast wastelands of Texas.
A: An intuitive way to think about it. Escaping an inverse square law to infinity requires energy (kinetic energy is this example) equivalent to the force of gravity times the radius. So the number od gees you would need is the ratio of track length divided by the radius of the earth. You have the additional complication that you must curve from horizontal velocity to velocity at and angle, and this means you are also dealing with centripetal forces during the transition from plain to mountain.
