The question is:
A ball, which is initailly at rest, is released at pointX and falls freely under gravity. It passes pointYand pointZ, which are 150m apart, as shown in the figure.
If it takes 5s for the ball toreach pointZ from pointY, how long dose it take to fall pointX to piontY?
http://img175.imageshack.us/img175/6238/aall2.th.png
The answer is:
s=ut+1/2(at^2)
150=u(5)+1/2(10)(5)^2
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要點樣才可以計出10出來.??
From.4-- physics!!
2008-03-16 9:22 pm
回答 (2)
2008-03-16 9:45 pm
✔ 最佳答案
Let g = 10m/s^2, let v = velocity at Y velocity at X = 0m/sFrom Y to Z,
s = ut +0.5gt^2
150 = v(5) + 0.5 (10) (5)^2
v= 5m/s
From X to Y
a = (v-u)/t
10 = (5 - 0) / t
t = 0.5s
2008-03-17 10:09 pm
When the ball is falling from X to Y,
uQ=0,vQ=?,a=10(take downward as positive),tQ=?
vQ=0+10tQ---(1)
When the ball is falling from Y to Z,
S=150m,tyz=5s,uyz=vQ,a=10
150=(vQ)5+(1/2)(10)(5)^2
so vQ=5
from1,,tQ=0.5s
uQ=0,vQ=?,a=10(take downward as positive),tQ=?
vQ=0+10tQ---(1)
When the ball is falling from Y to Z,
S=150m,tyz=5s,uyz=vQ,a=10
150=(vQ)5+(1/2)(10)(5)^2
so vQ=5
from1,,tQ=0.5s
收錄日期: 2021-04-14 18:58:02
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20080316000051KK01242