在无风的环境,某人在高处释放静止的篮球,篮球竖直下落;如果先让篮球以一定的角速度绕过球心的水平轴转动(如图)再释放,则篮球在向下掉落的过程中偏离竖直方向做曲线运动。其原因是,转动的篮球在运动过程中除受重力外,还受到空气施加的阻力
![](https://math.21cnjy.com/MathMLToImage?mml=%3Cmath+xmlns%3D%22http%3A%2F%2Fwww.w3.org%2F1998%2FMath%2FMathML%22%3E%3Cmrow%3E%3Cmsub%3E%3Cmi%3Ef%3C%2Fmi%3E%3Cmn%3E1%3C%2Fmn%3E%3C%2Fmsub%3E%3C%2Fmrow%3E%3C%2Fmath%3E)
和偏转力
![](https://math.21cnjy.com/MathMLToImage?mml=%3Cmath+xmlns%3D%22http%3A%2F%2Fwww.w3.org%2F1998%2FMath%2FMathML%22%3E%3Cmrow%3E%3Cmsub%3E%3Cmi%3Ef%3C%2Fmi%3E%3Cmn%3E2%3C%2Fmn%3E%3C%2Fmsub%3E%3C%2Fmrow%3E%3C%2Fmath%3E)
。这两个力与篮球速度
![](https://math.21cnjy.com/MathMLToImage?mml=%3Cmath+xmlns%3D%22http%3A%2F%2Fwww.w3.org%2F1998%2FMath%2FMathML%22%3E%3Cmi%3Ev%3C%2Fmi%3E%3C%2Fmath%3E)
的关系大致为:
![](https://math.21cnjy.com/MathMLToImage?mml=%3Cmath+xmlns%3D%22http%3A%2F%2Fwww.w3.org%2F1998%2FMath%2FMathML%22%3E%3Cmrow%3E%3Cmsub%3E%3Cmi%3Ef%3C%2Fmi%3E%3Cmn%3E1%3C%2Fmn%3E%3C%2Fmsub%3E%3Cmo%3E%3D%3C%2Fmo%3E%3Cmsub%3E%3Cmi%3Ek%3C%2Fmi%3E%3Cmn%3E1%3C%2Fmn%3E%3C%2Fmsub%3E%3Cmsup%3E%3Cmi%3Ev%3C%2Fmi%3E%3Cmn%3E2%3C%2Fmn%3E%3C%2Fmsup%3E%3C%2Fmrow%3E%3C%2Fmath%3E)
,方向与篮球运动方向相反;
![](https://math.21cnjy.com/MathMLToImage?mml=%3Cmath+xmlns%3D%22http%3A%2F%2Fwww.w3.org%2F1998%2FMath%2FMathML%22%3E%3Cmrow%3E%3Cmsub%3E%3Cmi%3Ef%3C%2Fmi%3E%3Cmn%3E2%3C%2Fmn%3E%3C%2Fmsub%3E%3Cmo%3E%3D%3C%2Fmo%3E%3Cmsub%3E%3Cmi%3Ek%3C%2Fmi%3E%3Cmn%3E2%3C%2Fmn%3E%3C%2Fmsub%3E%3Cmi%3Ev%3C%2Fmi%3E%3C%2Fmrow%3E%3C%2Fmath%3E)
,方向与篮球运动方向垂直。下列说法正确的是( )
![](http://tikupic.21cnjy.com/2022/07/11/8d/1c/8d1c4b9e9f8a3ca518890aabe415461b.png)
- A、
、
是与篮球转动角速度无关的常量 - B、篮球可回到原高度且角速度与释放时的角速度相同
- C、人站得足够高,落地前篮球有可能向上运动
- D、释放条件合适,篮球有可能在空中持续一段水平直线运动