题目内容
A、B两小球同时从距地面高为h=15m处的同一点抛出,初速度大小均为v0=10m/s.A球竖直向下抛出,B球水平抛出,(空气阻力不计,g取10m/s2).求:
(1)A球经多长时间落地?
(2)A球落地时,A、B两球间的距离是多少?
(1)A球经多长时间落地?
(2)A球落地时,A、B两球间的距离是多少?
(1)
(2)![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241333241971042.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824133324134377.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241333241971042.png)
(1)(5分)A球做竖直下抛运动,由平抛运动规律可知
①
②
③
联立方程①②③解得:
④
(2)(5分)B球做平抛运动,由平抛运动规律可知
⑤
⑥
联立方程②④⑤⑥解得:x=10m、y=5m
此时,A球与B球的距离L为:
所以,A球经1s落地;
A球落地时,A球与B球的距离![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241333241971042.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824133324259539.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824133324306649.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824133324368794.png)
联立方程①②③解得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824133324134377.png)
(2)(5分)B球做平抛运动,由平抛运动规律可知
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824133324478480.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824133324524652.png)
联立方程②④⑤⑥解得:x=10m、y=5m
此时,A球与B球的距离L为:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241333241971042.png)
A球落地时,A球与B球的距离
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241333241971042.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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