A science question for semi accurate yet fictional story idea concepts

Question

I am working on a fictional story, but I want to have a few accurate scientifically accurate (or as close as possible) situations. Question one is: if a person falling at full terminal velocity, or higher speeds, was somehow grabbed (by the torso region) and the grabber turned the falling person from a vertical angle to 45 positive degrees horizontal (opposite the side they were falling), could that redirection keep the falling person survive if they landed with a roll after the redirection.

(that's the best way to describe it, but to clarify, imagine someone falling straight down to a character's right, the standing character grabs the falling person on the right, redirects them toward the left at a horizontal angle and instead of going for a direct 90 degree shift the instead to a 45 degree (or any angle between 45 and 90) would the falling person be able to survive this sort of redirection?)

My guess, is that if the catcher could survive the obvious force of the falling person without his/her arms tearing off, that the person who got redirected (if done properly) should be able to survive the force as it would change from a sudden stop of great force to a slightly more controlled landing, and could be even more survivable if they rolled to a stop.

Answer 1

I'm afraid you'll have to come up with a different solution.

We can consider that the catcher at first brakes the fall before redirecting the person (pure redirection has already been considered by Eddy), but, although the situation is improved, it turns out that the forces involved are still too high.

From Torricelli's equation, a reduction in speed from $v_i$ to $v_f$ over a distance $\Delta x$ requires an acceleration $a$ (constant, in order to avoid peaks) with magnitude given by

$$ a_{\mathrm{brake}} = \frac{|v_f^2-v_i^2|}{2\Delta x}. $$

Pure redirection requires $a_c=v^2/r$, so, using $r\sim \Delta x \sim 1$m, we see that redirecting implies $a\approx v^2$, while for braking only half as much is needed ($a_{\mathrm{brake}}\sim v^2/2$), which is an improvement.

The free fall speed is about $55$ m/s and if we assume that a skillful enough person can roll to safety from $100\,$km/h $\approx 30\,$m/s, than the braking acceleration will be

$$a_{\mathrm{brake}}\approx\frac{55^2-30^2}{2}\approx 1000\,\mathrm{m/s}^2 \approx 100g,$$ while the redirecting acceleration that follows will be $a_c\approx 30^2=900\,\mathrm{m/s}^2 \approx 90g$.

While not as bad as pure redirection ($\sim 300g$), that still corresponds to too much of a force (especially when concentrated on relatively small areas of the body) for serious damage to be avoidable.

If you allow your hero to jump upwards and start braking the victim mid-air some $\Delta x\sim 10\,$m above the ground, then the required acceleration becomes $a_{\mathrm{brake}}\approx 10g$, which is more acceptable.

Answer 2

Let's suppose that the person is grabbed at a full arms length away from the 'catcher'. The eqation for centripetal acceleration (the acceleration experienced when moving along an arc of a circle) is $$ a=\frac{v^2} {r} $$ where $v$ is the speed of the object, in our case the terminal velocity of a human $v \approx 53m/s$, and $r$ the radius of the circle, in our case the length of an arm $r \approx 1m$. Thus $$ a \approx \frac{53^2} {1} m/s^2 = 2809 m/s^2 \approx 286g $$ where $g \approx 9.81 m/s^2$ is the acceleration due to gravity. For reference, an untrained human can black out from accelerations around $5g$ see here for details. Thus I would conclude that the 'catcher' would not be able to provide anything like enough force, either achieving nothing or, if they grabbed hold, loosing all their fingers. The faller would carry on moving as if nothing had happened. If the catcher was some kind of super hero that could provide enough force, the fallers body would be torn apart by the force, if grabbed by the torso probably loosing head, arms and legs, and possibly the torso would also disintegrate.

Edit: If the faller has a mass of $m \approx 80kg$ then the force required to perform this manoeuvre is $$ F = ma \approx 80 * 286g = 22880g $$ which is the weight of an object of mass $22880kg \approx 23 tonnes$. Thus the catcher would have to be able to support (ie pick up) such a mass with their arm, and the faller support such a mass with whatever part of their body they are caught by.