题目内容
从空中某处平抛一个物体,不计空气阻力,物体落地时末速度与水平方向的夹角为
=60°。取地面为零重力势能参考面,则物体抛出时其动能与重力势能之比为( )
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824125459293302.png)
A.1/3 | B.1/12 | C.3 | D.12 |
A
物体做平抛运动,假设落地速度为v,由于落地的速度方向与水平方向的夹角为α,故
水平分速度为
v0=vx=vcos![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824125459293302.png)
竖直分速度为
vy=vsin![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824125459293302.png)
由于平抛运动的水平分运动为匀速直线运动,故
v0=vx=vcos![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824125459293302.png)
由于平抛运动的竖直分运动为自由落体运动,故高度为h=![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824125459371947.png)
抛出时的动能为Ek0=
,抛出时的势能为Ep0=mgh=![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241254594801116.png)
因而动能与势能之比为cot2
=1/3
故选A.
水平分速度为
v0=vx=vcos
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824125459293302.png)
竖直分速度为
vy=vsin
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824125459293302.png)
由于平抛运动的水平分运动为匀速直线运动,故
v0=vx=vcos
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824125459293302.png)
由于平抛运动的竖直分运动为自由落体运动,故高度为h=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824125459371947.png)
抛出时的动能为Ek0=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824125459449975.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241254594801116.png)
因而动能与势能之比为cot2
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824125459293302.png)
故选A.
![](http://thumb2018.1010pic.com/images/loading.gif)
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如图所示,从地面上方某点,将一小球以7.5m/s的初速度沿水平方向抛出,小球经过1.0s落地。若不计空气阻力,取g =10m/s2,则可知小球抛出时离地面的高度为 m,小球落地时的速度大小为 m/s。
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241300132281614.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241300132281614.png)