物理小组的同学用如图所示的实验器材测定重力加速度，实验器材有：底座、带有标尺的竖直杆、光电门1和2组
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\u5728\u7b2c\u4e00\u4e2a\u6a21\u578b\u7684\u57fa\u7840\u4e4b\u4e0a\uff0c\u6765\u89e3\u4e00\u4e0b\u7b2c\u4e8c\u95ee\uff0e\u6ce8\u610f\u53bb\u5ba1\u9898\uff0e\u201c\u4ece\u7b2c\u4e00\u4e2a\u6c34\u6ef4\u79bb\u5f00\u6c34\u9f99\u5934\u5f00\u59cb\u8ba1\u65f6 \uff0c\u7b2cN \u4e2a\u6c34\u6ef4\u843d\u81f3\u76d8\u4e2d\uff0c\u5171\u7528\u65f6\u95f4T\u201d\uff08\u8fd9\u53e5\u8bdd\u5f88\u5173\u952e\uff01\uff01\uff01\uff09 \u5047\u8bbe\u7b2c\u4e09\u6ef4\u6c34\u843d\u81f3\u76d8\u4e2d\uff0c\u90a3\u4e48\u6709\u51e0\u6bb5\u65f6\u95f4\u95f4\u9694\u5462\uff1f\u7b80\u5355\u60f3\u4e00\u4e0b\uff0c\u662f\u4e0d\u662f4\u6bb5\uff1f\u90a3\u4e48\u7b2cN \u4e2a\u6c34\u6ef4\u843d\u81f3\u76d8\u4e2d\u5c31\u6709\uff08N+1\uff09\u4e2a\u65f6\u95f4\u6bb5\uff0c\u5373(N+1)t=T
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t=T/\uff08N+1\uff09
\uff081/2\uff09g\uff082t\uff09^2=H
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| v |
| g |
所以从开始释放到经过光电门1的时间t″=t′-t=
| v |
| g |
所以经过光电门1的速度v′=gt″=v-gt；
根据匀变速直线运动的推论得：两光电门间的距离h=
| v′+v |
| 2 |
| 1 |
| 2 |
（2）根据h=vt?
| 1 |
| 2 |
得，
| h |
| t |
| 1 |
| 2 |
则
| h |
| t |
| 1 |
| 2 |
所以重力加速度大小g=2k．
故答案为：（1）vt?
| 1 |
| 2 |
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