I first considered the forces that would be involved in a fall from the top of a two-story house.
ΔKE + ΔPE = 0 J
½m(vf2 – v02) + mg(hf – h0) = 0 J
½(70 kg)(vf2) + (70 kg)(9.8 m/s2)(0 m – 7.26 m) = 0 J
vf = 11.9 m/s
If we assume that the height of a two-story house is approximately 25 ft, or 7.26 m, the velocity of the person just before they hit the ground will be 11.9 m/s.
p = mv = (70 kg)(11.9 m/s) = 833 kg m/s
ΣF = Δp/Δt = (0 kgm/s + 833 kgm/s) / (0.001 s) = 8.33 x 105 N
By using momentum and up as positive, we calculate that the sum of the forces acting on the person when they stop to be 8.33 x 105 N. Assuming that this force is distributed over two legs, each leg would experience 4.17 x 105 N of force.
4.17 x 105 N / (π(.0127 m)2) = 8.22 x 108 N/m2
Assuming this force is distributed directly to the tibia, the force can be written as . Since the ultimate compressive strength of bone is 8.22 x 108 N/m2, the person will most likely suffer some injuries and end up in the hospital for the holidays.