Force on carabiner from free fall Suppose I weigh about $75~\text{kg}$ and I am attached to a (non-elastic) rope which is attached to a carabiner which can take at least $20~\text{kN}$.  Suppose furthermore that I reach terminal velocity before the rope is tight, a bit far fetched maybe, but pretend that I reach $530~\text{km/h}$.
What force is being applied to the carabiner once the rope finally is tight?
Since $530~\text{km/h}$ is slightly exaggerated, what speed should I expect to reach when free-falling $5$ meters while rock climbing?
 A: This is impossible to answer without a lot of additional information and assumptions. The force applied to the carabiner is, of course:
$$F = ma$$
Your mass is known, but the acceleration isn't. It depends on flexibility in the carabiner itself and in your body, and the linkage between the carabiner and your body. (Of course, we would normally include the rope also, but it is a given that this is inelastic.)
If we know how many seconds is required for you to stop, we can use the momentum-equals-impulse formula to get $f$, namely: $\Delta mv = F\Delta t$.  However, this $F$ will just be an average $F$ over the stopping time $\Delta t$, not an instantaneous $F$, and certainly not the maximum $F$, which is what we would like to know.
Over the course of the event, force is a function of time; it is not constant. The carabiner fails if the function, at any instant, that force exceeds its load handling.
A: As @Kaz suggests, the force being applied on you will be a function of time and will vary, so all you'll get from $m\Delta v=F\Delta t$ is the average force. But if you observe the force-time graph, it will look somewhat like this. 

In fact, it will be a lot steeper than this graph.
We can use that to approximate this graph to a triangle having its vertex at $F_{max}$ and its base of length $\Delta t$ on the time-axis.
The impulse, or the area under the graph equals
$$J=\int Fdt=\frac 12F_{max}\Delta t$$
We also know that the impulse equals the change in linear momentum given by $$J=m\Delta v$$
Equating the two we get, $$F_{max}=2m\frac{\Delta v}{\Delta t}$$
This $F_{max}$ will be the tensile force felt by the carabiner if we assume it to be massless, or almost massless compared to the person, in which case the string on the other end will also exert the same force $F_{max}$ on it. So if this $F_{max}$ exceeds $20kN$, the carabiner will break.
Also note that $\Delta t$ will have to be calculated experimentally, because I don't think there is a way to derive it using the data at hand.
