解:在直角$ \triangle A B C $中, 已知$ A B=2.5 m \text { , } B C=0.7 \mathrm{m} \text {, }$
则$ 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 $为斜边,
∴$C B_{1}=\sqrt{(A_{1} B_{1})^{2} +(C A_{1})^{2}} =1.5 \mathrm{m} $
∴$B B_{1}=C B_{1} +C B =1.5 + 0.7=0.8 \mathrm{m}$
$\text { 答: 梯足向外移动了 } 0.8 \mathrm{m} \text {. }$