Consider a scenario where a 100kg mass is hanging from a rope. It's stationary, so the force it exerts on the rope is given by $F=ma$, where $a$ is the acceleration due to gravity, which works out to be about 1000 N ($g = 10ms^2$ is an acceptable approximation for these purposes).
Let's instead consider a scenario where the same mass is in freefall, before suddenly being caught by the rope and coming to a stop. I'm using the following equations:
$h = \frac{1}{2}a t^2$
$v = at$
$F=ma = \frac{mv}{T}$
where $t$ is the time the object spends in freefall, and $T$ is the time it takes for it to stop. I'm not sure what's reasonable for the latter, as it's going to depend on some real-world factors like the stretch of the rope, etc. For now I'm going with $T=0.2$ s, and I can always adjust this later.
The first two equations can be rearranged into the form $v = \sqrt{2ah}$ to give the final velocity based on how far the object falls. Plug this into the third equation and I get: $F=\frac{m\sqrt{2ah}}{T}$
If I consider a very short fall of 10 cm, then this equation gives a value of about 700 N. This however is less than if the object where just sitting still and supported by the rope (1000 N). Intuitively I would have thought that if you give a solid yank on a rope it's going to exert more force than if you're just sitting there, so I'm clearly missing something.
Could someone explain where I've gone wrong, and how I should be handling this scenario?
