$解:∵AB⊥BH,CD⊥BH,EF⊥BH,$
$∴AB∥CD∥EF,$
$∴△CDG∽△ABG,△EFH∽△ABH,$
$∴\frac {CD}{AB}=\frac {DG}{BD+DG},$
$\frac {EF}{AB}=\frac {FH}{BD+DF+FH}.$
$∵CD=DG=EF=2米,DF=52米,FH=4米,$
$∴\frac {2}{AB}=\frac {2}{BD+2},\frac {2}{AB}=\frac {4}{BD+52+4},$
$∴\frac {2}{BD+2}=\frac {4}{BD+52+4},$
$解得BD=52,$
$∴AB=54米.$
$即建筑物的高为54米.$