物理 運動學
回答 (2)
2014-02-16 8:38 am
✔ 最佳答案
設石A,B,C開始下跌的地方是高度為 0，設初速度為 0。h = (V₀)(t) - (1/2)gt²
ha = (0)(t) - (1/2)gt²
ha = - (1/2)gt²
hb = - (1/2)g(t-1)²
hc = - (1/2)g(t-2)²
hb-hc = - (1/2)g[(t-1)² - (t-2)²]
= - (1/2)g[t²-2t+1-(t²-4t+4)]
= - (1/2)g[t²-2t+1-t²+4t-4]
= - (1/2)g[2t-3]
|hb-hc| = (1/2)g[2t-3]
B,C兩石相距 = (1/2)g[2t-3]
2014-03-06 7:10 am
收錄日期: 2021-04-27 20:52:37
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20140215000010KK04410