The volume $V$ of liquid in a spherical tank of radius $r$ depends on

the depht $h$ as follows:

$v=\frac{h^2(3r-h)}{3}$

Since $r=1$ and $V=0\mathrm{.}5m^3,$ calculate the $h$ value around $0\mathrm{.}5m$

with a tolerance of ${10}^{-7}$ using the 'simple iteration method: