Deflection of an Elastic Beam

When an elastic beam AB supports a block of weight W at a given point B, the deflection $$y_s$$ (static deflection) is proportional to W. Show that if the same block is dropped form a height h onto the end B of a cantilever beam, the maximum deflection $$y_m = y_s(1 + (1+\frac{2h}{y})^\frac{1}{2}).$$ Neglect the weight of the beam and any energy dissipated in the impact.

I have:

$$y_s = kW$$

T1 = 0
V1 = mgh
T2 = 0 when deflection is at a max
$$V2 = -mgy_m$$

I'm pretty sure that V2 should also include the potential energy stored in the beam, but I don't know how to express that. Would it be similar to a spring? V(beam) = 1/2ky^2 ? That's my guess, but there must be a logical way of proving it.

I think I can do the rest once I find the expression for the potential energy inside the elastic beam.

 PhysOrg.com science news on PhysOrg.com >> Hong Kong launches first electric taxis>> Morocco to harness the wind in energy hunt>> Galaxy's Ring of Fire
 Recognitions: Gold Member Homework Help Science Advisor Staff Emeritus Hey, check this out: http://www.engin.brown.edu/courses/En31/en31design.pdf

Recognitions:
Homework Help
 Originally posted by AngelofMusic I have: $$y_s = kW$$ ... ... there must be a logical way of proving it.
Yes, there is. You have identified a force law that should look familiar (if you put the k on the other side). How do you derive the potential energy from that force law (think about F dot dx).