$1.5.19.$ A spherical buoy of radius $R$ is attached to the bottom of the reservoir. The water level in the reservoir rises at a speed $u$. What is the speed of movement of the boundary of the flooded part of the buoy on its surface at the moment when the water level is $h$ above the center of the buoy?
By time $t$ the water level will be $h$
In a short period of time $dt$ the water layer will rise by
$$ dh = u \, dt $$
In turn, the boundary of the flooded part will also rise by
$$ dh = v\, dt \cos \alpha $$
Let's equate both equations
$$ v \cos \alpha = u $$
$$ v = \frac{u}{\cos \alpha} $$
$$ \fbox{$v = u \frac{R}{\sqrt{R^2 - h^2}}$} $$
$$v = u \frac{R}{\sqrt{R^2 - h^2}}$$