I have been working on this problem myself before giving it to my students.
My initial strategy was to reduce the problem to 10 floors at first. I tried beginning 'halfway' between (floor 5 or 6) and then continue this for every worst case scenario. However I thought that there would be a better strategy when the number of floors is increased. I tried other fractions such as starting a third of the way up but I was not able to generalise. After thinking about the problem some more for some reason the sum of the natural numbers came into my head. So I then decided to try an alternative strategy using the formulae for the sum of the natural numbers i.e. 1+2+3+...+n=n(n+1)/2 (=:S).
For 10 floors, let S=10 then solving for n we find n=4. This means that the best strategy is four drops. For example if we begin on the fourth floor and if the ornament does not break move to the seventh floor and if the ornament does not break move to the ninth floor and if the ornament does not break move to the tenth floor then we are done. If the ornament did break on the 4th, 7th, 9th or 10th floors then we work our way down one at a time until it does not break anymore. Therefore the maximum number of drops is 4 when we have 10 floors.
I then applied this for 100 floors i.e. S=100 then solving for n we find n =13.651 (positive root). Therefore the maximum number of drops is 14 when we have 100 floors. So in general, all that is necessary is to use the formula for the sum of the natural numbers and let this be equal to the number of floors. This is so simple and that's why I love this problem! I cannot wait to use it in a lesson!