1.An elevator under repair is descending at constant speed.A

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1.An elevator under repair is descending at constant speed.At the moment it passes the mechanic,he drops his wrench.The wrench hits the elevator after 1,4second.What is the speed of the elevator?What is the speed of the wrench at the moment of hitting?
2.For a freely falling body,what is the ratio of the time needed to cover the first half of the distance and the time needed to cover the second half of the distance?

俺不懂英语,汉语写中不?
1、撞击时,扳子速度为v0,则v0^2=1/2gt^2.可以解得v0.
由v0^2=2gh,可得扳子下落高度h,那么电梯速度v=h/t
2、设半程长为s,前半程时间为t1,后半程时间为t2,则：
1/2gt1^2=s
gt1t2+1/2gt2^2=s (s=vot+1/2at^2)
上两式可得：
t1^2-2t1t2=t2^2
利用求根公式可得t1=(2t2+sqrt(4t2^2+4t2^2))/2=t2+(sqrt2)t2
(sqrt表示开平方)
或t1=t2-(sqrt2)t2 (显然t1>t2,t1:t2一定大于一,所以这个舍去)
所以 t1:t2=1+sqrt2

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