-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 |