This begs the question for me - at terminal falling speed, what’s the fastest you can decelerate to 0 and not sustain injury? And given that, how much more distance would you need to move?
Maybe a superhero can catch you, decelerate you to 0 over 3 inches and that’s good enough?
Human terminal velocity is roughly 56 m/s. Let’s say our superhero wants to decelerate the person at 10G, which should be survivable for a short period. That would be 0.6 seconds of deceleration over 48 m. That’s a short time but quite a long distance, let’s slow down faster:
20G -> 0.28 seconds, 24 m.
30G -> 0.19 seconds, 16 m.
50G -> 0.11 seconds, 9.6 m.
100G -> 0.057 seconds, 4.79 m.
200G -> 0.029 seconds, 2.45 m.
5000G -> 0.0011 seconds, 3.6 inches.
A 40 mph car crash in a modern car into a solid wall gives around 15G.
F1 driver David Purley survived a 180G crash in 1977.
In short, I don’t recommend catching someone with 3 inches to spare.
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This begs the question for me - at terminal falling speed, what’s the fastest you can decelerate to 0 and not sustain injury? And given that, how much more distance would you need to move?
Maybe a superhero can catch you, decelerate you to 0 over 3 inches and that’s good enough?
Human terminal velocity is roughly 56 m/s. Let’s say our superhero wants to decelerate the person at 10G, which should be survivable for a short period. That would be 0.6 seconds of deceleration over 48 m. That’s a short time but quite a long distance, let’s slow down faster:
20G -> 0.28 seconds, 24 m.
30G -> 0.19 seconds, 16 m.
50G -> 0.11 seconds, 9.6 m.
100G -> 0.057 seconds, 4.79 m.
200G -> 0.029 seconds, 2.45 m.
5000G -> 0.0011 seconds, 3.6 inches.
A 40 mph car crash in a modern car into a solid wall gives around 15G.
F1 driver David Purley survived a 180G crash in 1977.
In short, I don’t recommend catching someone with 3 inches to spare.
Something is wrong with your distance formula by a factor of 4 (you should have D=1/2 * V_0^2 / a) . Not that it changes anything in your conclusions.
Wow. Good math. Thanks!