如图所示，从倾角为θ的斜面上的M点水平抛出一个小球，小球的初

问题多选题

如图所示，从倾角为θ的斜面上的M点水平抛出一个小球，小球的初速度为v₀，最后小球落在斜面上的N点，则（　　）

A．求出小球平抛运动过程的时间

B．不可能求出小球落到N点时速度的大小和方向

C．不可能求M、N之间的距离

D．可以断定，当小球速度方向与斜面平行时，小球与斜面间的距离最大

答案

设MN之间的距离为L，则由平抛运动的规律得

水平方向上：Lcosθ=V₀t

竖直方向上：Lsinθ=

gt²
由以上两个方程可以解得 L=

sinθ

gcos2θ

，

V₀tanθ，
所以A正确，C错误，

B、在竖直方向上，由自由落体的速度公式可得在N点时竖直速度的大小，

V_y=gt=g?

V₀tanθ=2V₀tanθ，
所以在N点时速度的大小为V=

tan

，

夹角的正切值为 tanβ=

=2tanθ，
所以B错误，

D、由物体的运动轨迹可以知道，物体离斜面的距离先变大在减小，当小球速度方向与斜面平行时，小球与斜面间的距离最大，所以D正确．

故选AD．

完形填空。
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