※ 引述《cutiehoneyzu (McHilo)》之銘言： : [領域] 運動學 : [來源] vector mechanics for engineers 7th : [題目]11.74 An elevator starts from rest and moves upward.accelerating at a : rateof 1.2m/s^2 until it reach a speed of 7.2m/s ,which it then maintains. : two seconds after the elevator begins to move ,a man standing 12m above : the initial position of the top of the elevator throws a ball upward : with an initial velocity 20m/s.Determine when the ball will hit the elevator : ^ 20m/s : o | : |--o ____ : _ _ _ ^_ : | | : | | : |a=1.2| 12m : | ^ | : | | | : |_ _ _| ____ : |eleva|tor : |_ _ _| : [瓶頸] 最困難的是在電梯向上加速後又變成等速上升 : y=v0t+0.5a t^2 又用不得因為它後來變為等速上升 : 而且人在2秒後向上拋出一球就變得很困難 (7.2)/(1.2)=6 elevator y= 0.5(1.2)t^2 + 7.2(t-6) ball y'=12 + 20(t-2) - 0.5(10)(t-2)^2 y=y' 應該是這樣吧@@ --

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