...I'm still highly skeptical that the lightweight plastic Glock trigger could ever do this...
The actual parts that will move and fire the gun in the absence of a trigger safety are the trigger bar and striker. The plastic trigger is just part of the trigger bar assembly which is mostly steel. The striker is a solid steel part.
Rather than let the speculation continue, I finally worked out the numbers. For those who want to make sure I'm not playing any games, here's the link to a calculator that will do the math for you. I'll provide the numbers in the right units to plug into the calculator.
http://hyperphysics.phy-astr.gsu.edu/hbase/flobi.html
The trigger bar and striker from a Glock 17 have a combined weight of 0.0188 kg (weighed on an RCBS electronic scale).
I assumed that the gun would be dropped from 4 feet=1.22 meters
When the gun hits the ground, it will be travelling about 16 feet per second and the striker and trigger bar combined will have an energy of about 0.221 Joules.
I assumed the gun would hit muzzle up and assumed it hit an unyielding surface and stopped instantly--did not bounce. In the real world, it would bounce and that would mean the force generated would be greater, but I didn't get into that because I didn't feel like trying to calculate what the bounce would be and because the calculator y'all are going to use to check my work doesn't do that either.
Based on my measurements from a Glock cutaway pistol, the striker and trigger bar will move about 6 mm between half cock and the point where the striker is released. That's 0.006 meters and that's the distance over which the kinetic energy will be dissipated. With that number the force generated by the impact can be calculated.
The calculator indicates that will generate 37 Newtons and you can use the Google conversion for Newtons to pounds force to validate that 37 Newtons is over 8lbs of force. The trigger on a Glock trigger can be operated by less than 7 lbs.
So, without a trigger safety, a Glock dropped from 4 feet and landing muzzle up on an unyielding surface would discharge.
Why wouldn't the firing pin stop it? Because
without a trigger safety the trigger bar will move along with the striker, disabling the firing pin safety in the process.
Because the trigger safety prevents the trigger bar from moving due to inertia/momentum, the passive safeties remain engaged. Even though the striker will bounce downwards on impact, when it returns forward it will be caught by the trigger bar, or, if that doesn't happen due to some bizarre malfunction, the firing pin safety will block it.
One function of the trigger safety is to insure that inertia/momentum of internal parts does not fire the gun when it is dropped. Just like Glock says.