Is the complexity of this algorithm O($\sqrt{n}$) or linear? Let's say I have two identical jars and I want to find the height that the jars will breaks when dropped from various heights. I can drop the jars from height increments using steps on a staircase. I want to solve this problem sublinearly.

I decided to increment using square numbers (1, 4, 9, 16..etc) for a complexity of O($\sqrt{n}$), where n is the number of steps on the stairs, so I know that I have an upper and lower bound once the first jar breaks. If the jar breaks at 16, I know that the height that it breaks at is between 9 + 1 and 16. If I were to run the algorithm linearly (on the second jar) from 10 to 16, would this make my algorithm linear?
No, it would not make the algorithm linear. The worst-case scenario would be $n = m^2 + 1$, where $m$ is some positive integer. In this case, you have an additional $(m+1)^2 - m^2 = 2m + 1$ many tries (since the next square after $m^2$ is $(m+1)^2$ and $n &lt; (m+1)^2$), which is roughly equal to $2\sqrt{n} \in \Theta(\sqrt{n})$.