题目内容
(12分)如图所示,跳台滑雪运动中,运动员经过一段加速滑行后从O点水平飞出,经过t=3.0s落到斜坡上的A点,已知O点是斜坡的起点,斜坡与水平面的夹角为θ=37°,运动员的质量为m=50kg,不计空气阻力,取sin37°=0.6,cos37°=0.8,g=10m/s2,求:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250012269422068.jpg)
⑴A点与O点间的距离L;
⑵运动员离开O点时的速率v0;
⑶运动员落到A点时的速度v。
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250012269422068.jpg)
⑴A点与O点间的距离L;
⑵运动员离开O点时的速率v0;
⑶运动员落到A点时的速度v。
⑴L=75m;⑵v0=20m/s;⑶v=
m/s,方向与水平方向的夹角为:α=arctanα![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001226974388.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001226958457.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001226974388.png)
试题分析:⑴运动员从O点水平飞出后,做平抛运动至A点,在竖直方向上,根据自由落体运动规律可知,运动员下落的高度为:h=
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001226989541.png)
根据图中几何关系可知:L=
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001227005554.png)
由①②式联立解得A点与O点间的距离为:L=
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001227036686.png)
⑵A点与O点间的水平距离为:x=
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001227036563.png)
在水平方向上运动员做匀速运动,所以有:v0=
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001227052368.png)
由①③④式联立解得运动员离开O点时的速率为:v0=
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001227067646.png)
⑶在竖直方向上vy=gt=30m/s
所以运动员落到A点时的速度大小为:v=
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001227083601.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001226958457.png)
设其方向与水平方向的夹角为α,则:tanα=
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001227114451.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001226974388.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001226974388.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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