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