如图所示,摩托车做腾跃特技表演,沿曲面冲上高0.8m顶部水平高台,接着以v=3m/s水平速度离开平台,落至地面时,恰能无碰撞地沿圆弧切线从A点切入光滑竖直圆弧轨道,并沿轨道下滑.A、B为圆弧两端点,其连线水平.已知圆弧半径为R=1.0m,人和车的总质量为180kg,特技表演的全过程中,阻力忽略不计.(计算中取g=10m/s2,sin53°=0.8,cos53°=0.6).求:                                                                                                                                     

(1)从平台飞出到A点,人和车运动的水平距离s.                                                    

(2)人和车运动到圆弧轨道最低点O速度v’=m/s此时对轨道的压力.                                      

(3)从平台飞出到达A点时速度及圆弧对应圆心角θ.                                                      

(4)人和车运动到达圆弧轨道A点时对轨道的压力.                                                         

                                                                       

                                                                                                                                     

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