-mgH-fH=0-
| 1 |
| 2 |
| v | 20 |
小球上升至离地高度h处时速度设为v1,由动能定理得:
-mgh-fh=
| 1 |
| 2 |
| v | 21 |
| 1 |
| 2 |
| v | 20 |
又由题有:
| 1 |
| 2 |
| v | 21 |
小球上升至最高点后又下降至离地高度h处时速度设为v2,此过程由动能定理得:
-mgh-f(2H-h)=
| 1 |
| 2 |
| v | 22 |
| 1 |
| 2 |
| v | 20 |
又由题有:2×
| 1 |
| 2 |
| v | 22 |
以上各式联立解得:h=
| 4H |
| 9 |
选项C正确,ABD错误.
故选:C
| H |
| 9 |
| 2H |
| 9 |
| 4H |
| 9 |
| 3H |
| 9 |
| 1 |
| 2 |
| v | 20 |
| 1 |
| 2 |
| v | 21 |
| 1 |
| 2 |
| v | 20 |
| 1 |
| 2 |
| v | 21 |
| 1 |
| 2 |
| v | 22 |
| 1 |
| 2 |
| v | 20 |
| 1 |
| 2 |
| v | 22 |
| 4H |
| 9 |