In the opening scene to the remade Bond film Casino Royal (2006), Bond chases a suspect who is played by the founder of freerunning Sebastien Foucan. In the scene he jumps from a crane to a second crane and then onto a building.

How much force does it take to stop his fall from the crane?
Solution: We can estimate the height of the crane above the roof from the size of the people at about 6.5 meters. Neglecting air resistance and using energy conservation we can find his velocity as he impacts the roof.

\frac{1}{2}mv^2 = mgh

v = \sqrt{2gh} = \sqrt{2(9.8m/s^2)(6.5m)} = 11.3 m/s

When either man lands on the roof their y momentum will go from mv to 0. If we take the force that causes this momentum change to be constant we have

\Delta p = \int Fdt = Ft \implies F = {\Delta p \over{t}}

Let’s say Bond’s mass is 80kg and he takes 0.1s to stop his fall. Then the force would be

F = {(80kg)(11.3m/s)\over 0.1s} = 9040N

Ouch! That’s going to hurt. Let’s say Foucan’s mass is 65kg and he takes 0.5s to stop his fall, then

F = {(65kg)(11.3m/s)\over 0.5s} = 1469N

Much better. This is why extending the time to stop the fall is a critical part of arts like freerunning.