# Freefall equations and forces

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?

Since your equations assume $$g=0$$ there is no reasonable reason why you force would have anything to do with the weight in the Earth's gravitational field, all you've calculated is the force require to stop an object in a vacuum.
All you need to do is add $$mg$$ and then you'll have the right answer for the scenario on Earth, that is,
$$F=mg+\frac{m\sqrt{2gh}}{T}$$.