For the 69 million lbs. to apply a force greater than 69 million lbs. requires a deceleration greater than 1g. The reality is with only 0.3g resistance the upper section continuously accelerated at 70% of gravity,
Pardon reality's intrusion: The 69 million pounds of mass already exerts a force of 69 million pounds even when it is at rest, because whatever is supporting it is already supplying sufficient force to counter 1g of gravitational acceleration.
If the 69 million pounds of mass is moving downward at .7g, then it will rapidly acquire momentum. After falling for, oh I don't know, let's say 10 feet at .7g, those 69 million pounds (about 31 million kilograms) will be travelling at about 21 feet per second (about 6.5 meters per second) and have about 200 million Newton-seconds of momentum. If you think the collision of those 69 million pounds with the floor below should bring them to rest, then the floor below will have to absorb the 200 million Newton-seconds of momentum with an opposing impulse before the floor breaks.
The floor is brittle and will break if it moves downward very far before the downward momentum is arrested. That means the downward momentum will have to be arrested quickly, before the floor moves far enough to break. For example, let's suppose the downward momentum is arrested within 65 milliseconds (mainly because that makes the acceleration come out to a round number). That requires an average deceleration of 100 meters per second squared for the 65 milliseconds, roughly 10g. The floor will deflect about 21 centimeters downward during those 65 milliseconds. It will also be exerting an upward force of about 3 billion Newtons during those 65 milliseconds. That's about 700 million pounds of force.
Looks like the floor's going to break. I don't know whether it's going to break from excessive force or from excessive deflection, but it's going to break.
How much of that 700 million pounds of force could the floor actually exert before it breaks? Some have said 3 million pounds, others 6 million; even if we accept your figure of 29 million pounds, the floor is going to exert only 4% of the required force before it breaks. How long will it take for the force on the floor to increase from zero to 29 million pounds? If the impact is perfectly solid, as gives the largest jolt, then the floor will break within a tiny fraction of 65 milliseconds, and the deceleration will never even get close to the 100 meters per second squared because the floor breaks at 4% of the force required to achieve that deceleration.
Just for grins, let's say it takes 10 milliseconds for the force to increase to 29 million pounds, and that the deceleration during those 10 milliseconds averages 4 meters per second squared. A deceleration of 4 meters per second squared, applied for 10 milliseconds, results in a delta-V of 4 centimeters per second.
So your missing jolt would be on the order of 4 centimeters per second, which is less than 1% of the 6.5 meters per second velocity. Are you certain a jolt of that magnitude would be observable on the videos?
The above, by the way, is what engineers mean by a back-of-the-envelope calculation. Maybe I made a mistake. You should consider running the numbers yourself sometime.
