试题与答案

一个物体从45m高处由静止自由下落,不计空气阻力.求: (1)前2s内的位移大小

题型:问答题

题目:

一个物体从45m高处由静止自由下落,不计空气阻力.求:

(1)前2s内的位移大小;

(2)最后1s内通过的高度及平均速度大小.

答案:

(1)根据位移时间关系公式,有:

S=

1
2
gt2=
1
2
×10×22m=20m.

(2)由h=

1
2
gt2代入数据得t=3s

故最后1秒内的位移为:

h′=45-

1
2
×10×22m=25m.

最后1秒内的平均速度:v=

h′
t
=25m/s.

答:(1)前2s内的位移大小为20m;

(2)最后1s内通过的高度为25m,平均速度大小为25m/s.

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