一物體自高於水面44.1m的橋上掉下，恰好落入以等速移動的船上，物體落下時，船離相遇點12m，求船速?
回答 (1)
2019-11-30 3:16 pm
用s=ut +(1/2)at^2
找物件掉落到船上的時間
u=0ms^-1
a=9.8ms^-2
s=44.1m
44.1 = (0)t+(1/2)(9.8)t^2
44.1 = 4.9t^2
t=0.9487s
再用s=ut+(1/2)at^2找船速
s=12m
a=0ms^-2
t=0.9487s
12=u(0.9487)+(1/2)(0)(0.9487)^2
u=12.6489ms^-1
有錯請賜教
找物件掉落到船上的時間
u=0ms^-1
a=9.8ms^-2
s=44.1m
44.1 = (0)t+(1/2)(9.8)t^2
44.1 = 4.9t^2
t=0.9487s
再用s=ut+(1/2)at^2找船速
s=12m
a=0ms^-2
t=0.9487s
12=u(0.9487)+(1/2)(0)(0.9487)^2
u=12.6489ms^-1
有錯請賜教
收錄日期: 2021-04-26 20:48:41
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20191129160755AAh468H