题目内容
如图所示,一物体自P点以初速度l0m/s做平抛运动,恰好垂直打到倾角为45°的斜面上的Q点(g=10m/s2)。则PQ两点间的距离为 ( )![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241252528081668.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241252528081668.png)
A.5m | B.l0m |
C.![]() | D.条件不足,无法求解 |
C
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241252528704541.png)
试题分析:小球垂直打到倾角为45°的斜面上,小球速度和竖直方向夹角为45°,此时速度可分解为竖直方向
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824125252886501.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824125252917600.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824125252932455.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824125252948638.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824125252995786.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824125253010843.png)
故选C
点评:平抛运动一般要分解成水平方向的匀速直线运动和竖直方向的自由落体运动来研究,应用平抛运动规律,找速度关系和位移关系来求解,尤其要注意角的应用。
![](http://thumb2018.1010pic.com/images/loading.gif)
练习册系列答案
相关题目