It is raining and a barrel of water has been filled to its maximum height of $90$ cm. Suddenly it stops raining, and the barrel of water is leaking water in such a manner that the level of water leaks with a speed that is proportional to the square root of the water depth.
How long does it take until the barrel is empty, if the level of water drops from $90$ cm to $85$ cm in an hour?
The correct answer is ~$ 35$ hours.
I try to put up a differential equation: $y'=\sqrt q y$, q: depth. Then I get uncertain about things, because $q$ varies (?), and I do not really understand why it helps me that the level of water sank $y'=5$ cm/h at one point? I have unmotivated guesses but do not really know how to continue, or if what I did first even is true in this case.
How would you tackle/interpret this problem and/or do you have any hints?
You seem to be confused by the wording of the question. As the other commenter pointed out, the depth of the water is the same as the level of the water.
Let $y$ be the water level, the the rate of change of the water is be given by $$ \frac{dy}{dt} = -k \sqrt{y} $$
where $k$ is some unknown constant, and the negative sign is to indicate the water level is decreasing.
You can solve the equation using separation of variables $$ \frac{dy}{2\sqrt{y}} = -\frac{k}{2} dt $$ $$ \sqrt{y} = C -\frac{kt}{2} $$ $$ y = \left(C - \frac{kt}{2} \right)^2 $$
Note that you have two unknown constants here. This is where the two given water levels come in. You have the initial water height is $y(0) = 90$ and the level after 1 hour is $y(1) = 85$
Plugging in those conditions, you get two equations $$ y(0) = C^2 = 90 $$ $$ y(1) = \left(C - \frac{k}{2}\right)^2 = 85 $$
Once you have the two constants, you'll need to find the value of $t$ so that $y=0$.