You're right about the velocity, but the impact force really depends on the properties of the material one lands on. First consider that our kinetic energy is

You're right about the velocity, but the impact force really depends on the properties of the material one lands on. First consider that our kinetic energy is K = mv^2/2 Additionally is the material has a spring constant k, then the force that results when the material deforms by x amount is F(x) = kx If we let d be the max distance the material, then we can integrate over the force to get the energy absorbed, which is equal to the kinetic energy. int_0^d F(x) dx = K kd^2/2 = mv^2/2 Now there are two ways the impact could work. Imagine we have a really thick crash pad, then we consider it to deform without limit and have a constant k, giving d = sqrt(mv^2/k) F_max = F(d) = v sqrt(mk) But what about a helmet? It can only over it's thickness, at which point your head is basically in contact with the concrete. Thus, we have a constant deformation distance d, and get k = mv^2/d^2 F(d) = mv^2/d As we can see, depending on how we consider it we get either a force of sqrt(10) or 10 times the original amount. I probably failed to take into account something that someone who knew more about materials could point out. tl;dr it's complicated because impact force does not necessarily scale linearly with velocity. update: so after some conversation with others. It seems that head on concrete is better modeled by a constant d. So a drop of 10 times the height results in 10 times the force. See also http://hyperphysics.phy-astr.gsu.edu/hbase/impulse.html http://hyperphysics.phy-astr.gsu.edu/hbase/impulse.html