| 1 |
| 2 |
| 1 |
| 2 |
得:t=
| H |
| v0 |
由题意:两物相遇时,v=v0-gt,2v=gt…③
得:v0=
| 3 |
| 2 |
由③④得:
| v | 20 |
| 3 |
| 2 |
则相遇地点离地面的高度为:h=v0t-
| 1 |
| 2 |
| H |
| v0 |
| 1 |
| 2 |
| H |
| v0 |
由⑤⑥得:h=
| 2 |
| 3 |
B上升的最大高度为:hmax=
| ||
| 2g |
由⑤⑧得:hmax=
| 3 |
| 4 |
由2v=gt,v0=
| 3 |
| 2 |
得:v0=3v…⑩
设A落地速度大小为vA.则:vA=
| 2gH |
由⑩代入得:vA=
2
| ||
| 3 |
| 3 |
故选:ABC
| 2H |
| 3 |
| 3H |
| 4 |
| 1 |
| 2 |
| 1 |
| 2 |
| H |
| v0 |
| 3 |
| 2 |
| v | 20 |
| 3 |
| 2 |
| 1 |
| 2 |
| H |
| v0 |
| 1 |
| 2 |
| H |
| v0 |
| 2 |
| 3 |
| ||
| 2g |
| 3 |
| 4 |
| 3 |
| 2 |
| 2gH |
2
| ||
| 3 |
| 3 |