题目内容
证明:任意时刻速度的反向延长线一定经过此时沿抛出方向水平总位移的中点
见解析
证:平抛运动示意如图
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241200208852702.jpg)
设初速度为V0,某时刻运动到A点,位置坐标为(x,y ),所用时间为t.此时速度与水平方向的夹角为
,速度的反向延长线与水平轴的交点为
,位移与水平方向夹角为
.
依平抛规律有:速度: Vx= V0
Vy=gt
①![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120021056169.png)
位移: Sx= Vot
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120021072641.png)
②
由①②得:
即
③
所以:
④
④式说明:做平抛运动的物体,任意时刻速度的反向延长线一定经过此时沿抛出方向水总位移的中点。
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241200208852702.jpg)
设初速度为V0,某时刻运动到A点,位置坐标为(x,y ),所用时间为t.此时速度与水平方向的夹角为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120020900339.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120020947291.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120020978310.png)
依平抛规律有:速度: Vx= V0
Vy=gt
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241200210251137.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120021056169.png)
位移: Sx= Vot
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120021072641.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241200211031188.png)
由①②得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120021181747.png)
即
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120021306779.png)
所以:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824120021353493.png)
④式说明:做平抛运动的物体,任意时刻速度的反向延长线一定经过此时沿抛出方向水总位移的中点。
![](http://thumb2018.1010pic.com/images/loading.gif)
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