Solution: The greatest number of heights that could be tested with certainty is 91.
If only one vase was available, we would have to begin at 1 foot and proceed upwards by 1 foot increments. If we skipped a height and the vase broke, we could not be certain that the maximum survival height was the last drop survived or a height which was skipped. If and when the first vase breaks, the problem will become a one-vase-scenario.
If two vases are available, we do not at first need to drop from consecutive heights, but we must choose a strategy which allows us to test the intervening heights between the first failure and the previous success.
With two vases, therefore, we can spend our first drop at 13 feet, knowing that we have 12 drops left with which to determine (in case of a failure), which of the heights 1-12 was the maximum. If the vase sustains a drop of 13 feet, we have 12 drops left and we should make our next drop from 25 feet, knowing that we will have 11 drops (in case of a failure) with which to determine the maximum that may be anywhere from 14 to 24.
Therefore, we can skip the first twelve heights, testing at 13. Then we skip the next eleven, testing at 25. By continuing this method, we can determine with certainty the safety of 91 different heights.
We plan to test at:
25 ft. (13+12)
36 ft. (25+11)
46 ft. (36+10)
55 ft. (46+9)
63 ft. (55+8)
70 ft. (63+7)
76 ft. (70+6)
81 ft. (76+5)
85 ft. (81+4)
88 ft. (85+3)
90 ft. (88+2)
91 ft. (90+1)
The winning student discussed 3, 4, and 5 vases. There would be a maximum, of course, if you had 14 vases and 13 drops, you would not benefit from the fourteenth vase.
Robert suspected a twist (in the comments for the question post) such as counting the first part of the fall. For example, if you drop a vase from 98 feet, it may smash at the end but it will safely fall 97 feet. However, a drop from 97 feet may also be fatal (in which case you have spent both of your vases and very little has been learned).