A small block slides down from the top of hemisphere of radius R =...

$mgcos\theta = \dfrac{m{v}^{2}}{R} . . . ( 1 )$

$cos\theta = \dfrac{h}{R}$

Energy conservation

$mg { R - h } = \dfrac{1}{2}m{v}^{2}$ $. . . ( 2 )$

from $( 1 )$ and $( 2 ) ⇒ mg\left\{\dfrac{h}{R}\right\} = \dfrac{2mg { R - h }}{R}$

$h = \dfrac{2R}{3} = 2m$