3 invaluable equations of motion:-

V^2=U^2 + 2AS
V=U + AT
S=UT + 1/2 A * T^2

For this one, we can get away with the first equation.

Let's imagine there is no atmosphere. No Drag.

V^2=U^2 + 2AS
We can equate velocity (U) at the surface of the Earth required to reach 100km altitude (S), where the parabola tops out and (V)=0. Acceleration (A) =1 g (the difference at 100 km altitude is minor).

Without needing to solve it, we can find how long the launcher needs to be. We need to acheive the same U, with a different A and S. Note that the equation resolves to A*S=constant

Suppose the launcher subjected the passengers to 5g (pretty rough for tourists, but possibly survivable). In this case, the launcher needs to be 20 km long.

if you can cope with 10g, then a "mere" 10 km is required.

I think you can see that the launch is not survivable from any tower, tunnel, or pit we can dig which is vertical. If you do it horizontally, you need a bit more speed, and have to also allow for the curve to vertical flight. This is pretty much what a rocket plane does. It takes off horizontally, flies to a fair height, then drops from the carrier plane, turns vertical and WHOOSH, a 2-3g pull for 20-50 km height...