题目内容
如图所示,在“研究平抛物体运动”的实验中,用一张印有小方格的纸记录轨迹,小方格的边长L=1.25cm。由于某种原因,只拍到了部分方格背景及小球在平抛运动途中的几个位置如图中的a、b、c、d所示,则小球平抛的初速度表达式v= (用L、g表示),大小是 m/s。(取g=9.8m/s2)![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824172256260284.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824172256260284.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824172256494490.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824172256712676.png)
试题分析:相邻两点间的时间间隔为T竖直方向:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824172257024610.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824172257367608.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824172257945955.png)
代入数据解得
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824172256712676.png)
点评:题是频闪照片问题,频闪照相每隔一定时间拍一次相,关键是抓住竖直方向自由落体运动的特点,由
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824172258740559.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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