Originally Posted by
B2N2
There is some flawed thinking here.
An object released to freefall will experience a 1G acceleration until atmospheric friction/drag will stabilize its speed at terminal velocity.
Depending on surface area and drag coefficient this terminal velocity may be significantly less then a dive while still under engine power.
https://www.grc.nasa.gov/www/k-12/airplane/termv.html
Recovery attempt from a steep dive can cause very high G-forces on the airframe
An example of this is here where they descended under power at appr. 12000fpm (!)
https://en.wikipedia.org/wiki/China_Airlines_Flight_006
The fact that the time of freefall more or less equals the time to descent is nothing but coincidence.
No, you are not correct here.
As I am sitting here completely at stand still on my chair in front of my computer I am experiencing exactly 1G (in the vertical direction), the same 1G as the unfortunate pilots in this flight experienced while they briefly was semi stable at 6000 feet. If a 6000 feet hole suddenly appeared under my chair I would instantly start free falling at zero G. If I was to reach the bottom of the hole in 20 sec I would need to fall at an average zero (vertical) G. The acceleration you are describing does not count as a positive 1G. An object which is free falling is by definition experiencing zero G, this is according to laws set by Albert Einstein.