解：如圖，連接?$ON、$??$OB.$?
∵?$OC\perp AB，$?
∴?$D$?為?$AB$?的中點(diǎn)?$.$?
∵?$AB = 7.2m，$?
∴?$BD=\frac {1}{2}AB = 3.6m.$?
設(shè)?$OB = OC = ON = rm，$?
則?$OD=(r - 2.4)m.$?
在?$Rt\triangle BOD$?中，根據(jù)勾股定理，得?$OB^2=OD^2+BD^2，$?
即?$r^2=(r - 2.4)^2+3.6^2，$?
?$ $?展開式子得?$r^2=r^2-4.8r + 5.76+12.96，$?
?$ $?移項(xiàng)可得?$4.8r=18.72，$?
?$ $?解得?$r = 3.9.$?
∵?$CD = 2.4m，$?船艙頂部高出水面?$AB 2m，$?
∴?$CE=2.4 - 2 = 0.4(\mathrm {m})，$?
∴?$OE = 3.9 - 0.4 = 3.5(\mathrm {m}).$?
易知?$OC\perp MN，$?
∴?$MN = 2EN.$?
在?$Rt\triangle OEN$?中，?$EN=\sqrt {ON^2-OE^2}=\sqrt {3.9^2-3.5^2}=\sqrt {15.21 - 12.25}=\sqrt {2.96}(\mathrm {m}).$?
∴?$MN = 2EN = 2×\sqrt {2.96}≈3.44(\mathrm {m}).$?
∵?$3.44＞3，$?
∴此貨船能順利通過(guò)這座拱橋
