4 ms·
"Turning momentum against the wall" needs actually more force than just stopping the faller. If the faller initially falls with momentum -p z (where I will deno
by yk 8y ago
"Turning momentum against the wall" needs actually more force than just stopping the faller. If the faller initially falls with momentum -p z (where I will denote the unit vectors of the coordinate system with x, z, and p is the magnitude of the momentum) and turns that into p x, then you need an average force of Favg=deltap /t= sqrt(2)p /t, while just stopping would only be Favg=p/t, where deltap is the total change in momentum and t is the duration of the stopping.
The reason for that is, that momentum is a vector and so you need to supply force in along each of the coordinate axis individually.
- LorenPechtel 8y agoYou're assuming a 90 degree turn. It doesn't take a 90 degree turn, however--they were probably still heading mostly down when they struck the wall. The wall (and the falling person) would absorb most of the impact. Not good for the faller but far better than keeping on going.