$解:過A作AE⊥OM于E,過B作BF⊥OM于F$
$∵∠AOE+∠OAE=90°$
$∠AOE+∠FOB=90°$
$∴∠OAE=∠FOB$
$在△AOE和△OBF中$
${{\begin{cases} {{∠AEO=∠OFB}} \\ {∠OAE=∠BOF} \\ {AO=OB} \end{cases}}}$
$∴△AOE≌△OBF(AAS)$
$∴OE=BF,AE=OF$
$∴OE+OF=BF+AE=17 m$
$∴EF=EM-FM=AC-BD=7 m$
$∴OE=\frac {1}{2}(OE+OF-EF)=5 m$
$AE=OF=12 m$
$∴ON=OA=\sqrt {AE^{2}+OE^{2}}=13 m$
$∴MN=OM-ON=OF+FM-ON=2 m$
$答:離地最低點(diǎn)高度MN為2m.$