题目内容
某同学在描绘平抛运动轨迹时,得到的部分轨迹曲线如图所示。在曲线上取A、B、C三个点,测量得到A、B、C三点间竖直距离
=10.20cm,
=20.20cm,A、B、C三点间水平距离
=12.40cm,g取10m/s2,则物体平抛运动的初速度大小为______m/s,轨迹上B点的瞬时速度大小为________m/s。(计算结果保留三位有效数字)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250005572074496.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000557113340.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000557145361.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000557160412.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250005572074496.png)
1.24m/s 1.96m/s
试题分析:据题意,由于物体做平抛运动,则有h2-h1=gt2,可以求出物体在AB、BC段的运动时间为t=0.1s,则物体做平抛运动的初速度为v0=x/t=1.24m/s;轨迹上B点的速度为vB=
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825000557223540.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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