如图，人的身高CG=EH=1.7m，路灯高为AB，第一次影子长为CD=1.3m，第二次影子长为EF=1.8m，2s内人前进的距离CE=1m/s×2s=2m，根据题意得：
（1）△GCD∽△ABD，∴
| CG |
| AB |
| CD |
| BC+CD |
| 1.7m |
| AB |
| 1.3m |
| BC+1.3m |
（2）△HEF∽△ABF，∴
| HE |
| AB |
| EF |
| BC+CE+EF |
| 1.7m |
| AB |
| 1.8m |
| BC+2m+1.8m |
（1）（2）两式联立解得：BC=5.2m，AB=8.5m．
故答案为：8.5．
| CG |
| AB |
| CD |
| BC+CD |
| 1.7m |
| AB |
| 1.3m |
| BC+1.3m |
| HE |
| AB |
| EF |
| BC+CE+EF |
| 1.7m |
| AB |
| 1.8m |
| BC+2m+1.8m |