2U Question. Halp.
Just to extend on Rui's answer (I get the same outcome), you could consider it in a similar way. This is definitely 2U content, but a really weird one, they like to do strange stuff with Pythag for some reason
it could show up in an Extension exam I suppose, but I see it more as a Band 6 2U question
The string is wrapped around a cylinder right, so what we can almost do (bear with me) is assign a
cylindrical coordinate system. This sounds complicated, but it's actually easy if we think about it practically.
Picture grabbing the end of the string at the top and unravelling it from the cylinder by pulling directly outward, not changing it's height, and leaving the other end fixed on the ground. This forms a right angled triangle where the string is the hypotenuse (and hence comes back in Rui's method, but I'm considering the whole cylinder at once).
The height of this triangle is still the same as the height of the cylinder (remember I said don't change the height), so that is h. The base is whatever distance the string travelled around the
circumference. What I've almost done is initially considered the x (horizontal) axis as wrapped around the cylinder, and then when I unravel it, I've moved back to our regular Cartesian system. But the distance is the same, that string has wrapped itself around that cylinder twice, so the 'horizontal' distance travelled is double the circumference. I've taken the horizontal distance as measured
around the cylinder (double the circumference), and stretched it out into a straight line.
Then, as Rui said, we use Pythagoras:
To simplify all that, I copied Rui's method but considered the whole cylinder at once, just to make it more practical. You COULD consider it in terms of changing coordinate systems, but that's unnecessary if you can picture unravelling the string. I threw it in to make myself sound smart