Gun Launchers
Postulate some magic gun that imparts velocity to a long, skinny, dense projectile, which subsequently pushes through the atmosphere on the way to space. What are the physics here?
Euler's Column Formula
The first thing to know is that long, skinny things loaded at one end want to buckle. If a columnar projectile is pushed at one end, and free at the other, and free to flex, the ends will push together and the column may possibly bow to one side. The model is equivalent to half the force pushing inwards at each end, compressing the column, with an additional force pushing on the middle.
With both ends free to pivot (unfortunately), Euler's formula is
F = \pi^2 E I / L^2
F 
force on the ends 
half the acceleration force 
E 
modulus of the structural material 

I 
moment of inertia 
\pi t r^3 / 16 for tube 
\rho_S 
density of the structural material 

L 
length of the tube 

t 
tube wall thickness 

M 
total projectile mass 

K 
structural mass fraction 

V 
velocity 

Vol 
payload volume 

B 
launch path length 

r 
radius of tube 

a_1 
acceleration in the launch tube 

a_2 
deceleration in air 

C_D 
drag coefficient 

\rho_A 
density of air 

\rho_P 
density of the payload 

H 
scale height of atmosphere, about 10 Km 

R_E 
radius of the earth 
All units MKS.
Assume the projectile is a round thick tube of something strong containing something too weak to contribute much flexural strength. Things are more complicated than that, but the complications do not improve the situation. For example, panel buckling on a thin shell will occur at a fraction of the expected load on a panel that stays flat. Nose cones add mass, and must be very strong and temperature resistant ( a sharp point will break off, for example).
The total tube mass is K M = 2 \pi r t L \rho_S , so t = K M / ( \pi r L \rho .
E I = ( E / \rho_S ) K M r^2 / 16 L and F = M a = \pi^2 ( E / \rho_S ) K M r^2 / 16 L^3 .
The maximum acceleration is a < \pi^2 ( E / \rho_S ) K r^2 / 16 L^3 .
The payload volume of the tube is Vol = \pi r^2 L
This works out to a < \pi ( E / \rho_S ) K Vol / 16 L^4 , or L^4 < \pi ( E / \rho_S ) K Vol / 16 a
For a given volume, we want something short and wide to resist buckling. But we also want something long and narrow to punch through the atmosphere.
Hmmm.
Assume maximum strength Zylon, with E = 5.8e9 Pa and \rho_S = 1500 kg/m^{3}. In Real Life&tm; you will probably derate that by a factor of 4 for a filament wound shell in a matrix, but lets assume best possible case. Thus E / \rho_S = 3.87e6 m^{2}/s^2 $.
The drag coeefficient C_D of a long, pointy nosecone is about 0.3 subsonic  not sure what happens hypersonic, but chances are it gets worse.
Air is front of the nose cone is compressed (a LOT) so that it turns into a ball of incandescent plasma , and explodes to the sides. But let's ignore the detailed behavior, and just assume that the deceleration in air is M a_2 = \rho_A V^2 \pi r^2
What is the deceleration path through dense atmosphere? If we are firing horizontally, it is proportional to \sqrt{ 2 R_E H } , about 360km.
We want to maximize the horizontal velocity remaining when we reach the vacuum of space.
If we fire straight up, it is 10 km, but this imparts no horizontal velocity to the payload, which is what we need to achieve orbit, rather than plunge straight back to earth. If we need to add all the horizontal velocity, then we need the same amount of rocket \Delta V that we would need from a conventional launch, and the projectile gun is just an added barrier to launch, making the rocket structurally much heavier without adding useful orbital velocity.
So, lets assume a 45 degree launch angle. That means we are punching though \sqrt{2} times the scale height of the atmosphere at tube exit altitude. That is about 13 kilometers at 5000m above sea level.
Assume we want to produce at least half of orbital velocity from our launch, which means a horizontal component of 4km/s, and a total speed of about 6km/s. Our long skinny projectile will see a drag force of \pi r^2 C_D \rho_a V^2 , half of which is compressing our column at both ends.
So 0.5 \pi r^2 C_D \rho_a V^2 < \pi^2 ( E / \rho_S ) K M r^2 / 16 L^3 .
Solving for maximum L, L^3 < \pi ( E / \rho_S ) K M / ( 8 C_D \rho_a V^2 ) .
Let's throw some numbers at this. If K = 0.5 and M = 10,000 kg, C_D = 0.3, \rho_A = 0.6 kg/m^{3} (6000m altitude) and E / \rho_S = 3.87e6 m^{2}/s^{2}, then L^{3} < 3000 m^{3} and L < 7 m .
If M = 200 kg, then L < 2m . Otherwise, our payload will buckle when it hits atmosphere.
LOX and RP1 have densities around 1000 kg/m^{3}; solid propellant is about twice that, though a real solid motor has a starcrosssection hole through it, so it also averages around 1000 kg/m^{3}.
The volume of (1K) M = 5000 kg is 5 m^{3}, and if L = 7 meters then r = 0.48 meters. The drag is 4.6 million newtons, the deceleration is 460 m/s^{2}. We are traversing 13 kilometers at 6 km per second, so that is a velocity loss of 1000 meters per second. Rinse and repeat  to leave the atmosphere at 6000 meters per second, we must exit the tube at around 7100 m/s, average velocity 6550 m/s, average drag 550 m/s^{2} for 2s.
Except ...
We also can't buckle in the tube, and the tube length is limited. The highest mountains are on high plateaus, so a launch tube running up their sides doesn't start at sea level, but some distance above. Unless we are burrowing deep into that plateau (can you say "geothermal heat?") we have a limited vertical run, so with the same limited acceleration set by the column formula, we can only exit the tube so fast. Kilimanjaro starts from a base of 900 meters; our mountainside launch tube is at best 5000 meters vertical, 7000 meters length. So to reach 7100 m/s, we need an acceleration of a_1 3600 m/s^{2}. L^4 < \pi ( E / \rho_S ) K Vol / 16 a , or L < 4.8m . That means r = 0.58 m, the average deceleration is 660 m/s^{2}, we have to launch faster, and ...
Unless the projectiles are above about 20 tons, they lose too much velocity on the way through the atmosphere. Achieving full horizontal orbital velocity with a higher speed launch, higher acceleration launch is probably off the table until projectiles weigh more than 100 tons. A gas based launcher will be accelerating a lot more gas than it accelerates payload, so peak energy levels in these systems is on the order of 1/2 M_gas_ V^{2}, (f M_gas_ = 50 tons and V_gas_ = 7000 m/s, then total energy is 1.2 Terajoules, which must thermalize and dissipate through the sides of the launch tube. Without the help of Maxwell's demon, that energy is lost, and probably melts the system.
This is kind of pointless, we still need a hell of a big rocket to add the other 4km/s. All we've done is got rid of the first stage, at the expense of a heck of a lot of expensive structural material for two or three more upper stages that no have to survive 50 times the stress of a rocket boost first stage. A high I_{SP} liquid fueled rocket will be too fragile  current rockets have a hard enough time launching from zero velocity, and are built lightly enough that turbulence at max Q at low mach and much higher altitude is still a problem. For a high gee launcher, we are stuck with lower I_{SP} solids for the entire stack.
Perhaps we can launch liquid fuels if they are frozen solid, for later use, but they will not be thawed when we need \Delta V at the top of our ballistic trajectory. Remember that even solids compress and squash sideways (Poisson's ratio) so we will need to contain them inside our tube, adding to the strength it needs.
BTW, you better have a way to keep that projectile pointed absolutely straight and centered in the tube. If it skids along the side of the tube  game over.
Also, if this is a gas gun, some of what is behind the projectile will leak into the front of it. The projectile is, after all, moving no slower than the gas layer pushing it, and if that is a partially thermalized gas, some of the molecules will go much faster than average. The vacuum in front of the projectile will be a lousy vacuum pretty soon. If the gas pressure behind is enough to accelerate a 10 ton projectile at 3600 m/s^{2}, the force is 36 MN, the pressure is around 400 atmospheres. A significant fraction of that will leak past the side of the projectile, depending on how close the projectile comes to the sides of the chamber.
During WW1, the German "Paris Gun" suffered so much barrel erosion that each shell fired was designed to have a slightly larger diameter, increasing in size until the barrel was replaced. Imagine the logistics nightmare of keeping this monster supplied; one missing batch of intermediate size, and the rest of the series was useless, change the barrel and start again. But then, the Wehrmacht was pretty idiotic about their logistics, and gas dynamics limited the performance of the Paris Gun, fortunately for the world.
Like many ideas, including most kinds of electromagnetic launcher, you must look for the worst case before nature finds it for you. People have been making guns for a thousand years  if there was any low hanging fruit, it would have been discovered long ago. We can certainly optimize their performance with modern materials, precision machining, computation and radar, but Newton's laws and gas dynamics still apply.