ssssssssss物理
回答 (2)
2010-03-27 12:50 am
✔ 最佳答案
from begining to the maximum point :v=u+at =>0=10+(-10)t =>t=1 s
so the sand bag reach the maximum after 1 s
and the sand bag back to the starting point after 1 s ( total 2 s )
also , now the bag is 4 10 ms^-1
s=ut+(1/2)at^2=(10)(2)+(1/2)(10)(2)^2=20+20=40m//
2010-03-27 3:54 am
You could use the equation of motion directly: s = ut + (1/2)at^2
with u = 10 m/s,t = 4 s, a = g (= -10 m/s2), s = ?
hence, s = 10 x 4 + (1/2)(-10)(4x4) m = -40 m
Therefore, the height of the balloon is 40 m above ground.
with u = 10 m/s,t = 4 s, a = g (= -10 m/s2), s = ?
hence, s = 10 x 4 + (1/2)(-10)(4x4) m = -40 m
Therefore, the height of the balloon is 40 m above ground.
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原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20100326000051KK00844