题目内容
A球由塔顶自由落下,当落下a米时,B球自距塔顶b米处开始自由落下,两球恰好同时落地,则A球下落的时间比B球下落的时间长 ,塔高 。
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241300497002731.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241300497163039.png)
分析:A球下落高度a的时间即为A球下落的时间比B球下落的时间长的时间,分别对AB两球运用自有落体位移时间公式即可解题.
解答:解:根据h=
gt2得:t=![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824130050090553.png)
设塔高h,则b球下落的时间为:tb=![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824130050090553.png)
对a球有:h=
g(t+tb)2
由①②解得:h=![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824130050200663.png)
故答案为:
,
,
解答:解:根据h=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824130049747338.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824130050090553.png)
设塔高h,则b球下落的时间为:tb=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824130050090553.png)
对a球有:h=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824130049747338.png)
由①②解得:h=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824130050200663.png)
故答案为:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824130050090553.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824130050200663.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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