解:在直角?$ \triangle A B C $?中, 已知?$ A B=2.5 m \text { , } B C=0.7 \mathrm{m} \text {, }$?
則?$ A C=\sqrt {2.5^2+ 0.7^2}=2.4\ \mathrm {m} ,$?
∵?$A C=A A_1+C A_1, $?
∴?$C A_1=2\ \mathrm {m},$?
∵ 在直角?$ \triangle A_1\ \mathrm {B}_1\ \mathrm {C} $?中?$, A B=A_1\ \mathrm {B}_1 , $?且?$ A_1\ \mathrm {B}_1 $?為斜邊,
∴?$C B_{1}=\sqrt{(A_{1} B_{1})^{2} +(C A_{1})^{2}} =1.5 \mathrm{m} $?
∴?$B B_{1}=C B_{1} +C B =1.5 + 0.7=0.8 \mathrm{m}$?
?$\text { 答: 梯足向外移動(dòng)了 } 0.8 \mathrm{m} \text {. }$?