$?解：過點P作PG⊥EF ，垂足為點G ，如圖所示?$
$?由題意得， AB//CD//PG?$
$?因為AB//PG?$
$?所以△PGD∽△ABD?$
$?所以\frac {PG}{AB}=\frac {DG}{DB}?$
$?因為CD//PG?$
$?所以△PGB∽△CDB，?$
$?所以\frac {PG}{CD}=\frac {GB}{DB}?$
$?因為\frac {DG}{DB}+\frac {GB}{DB}=\frac {DB}{DB}=1?$
$?所以\frac {PG}{AB}+\frac {PG}{CD}=1?$
$?因為AB=4m，CD=6m?$
$?所以\frac {PG}{4}+\frac {PG}{6}=1?$
$?所以PG=\frac {12}{5}m?$
$?答：交點P離地面的高度為\frac {12}{5}m$
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