Re: [問題] 斜拋位移持續增加的角度值
提供不用微積分的解法
思路:位移持續增加,代表全程位移和速度均夾銳角。
列出內積後,判別式小於零
https://i.imgur.com/8QJXCiT.jpg
![](https://i.imgur.com/8QJXCiT.jpg)
![](https://i.imgur.com/ffKY52w.jpg)
: : 如圖
: : 假設一物體以相同初速斜向拋出,若該物與原點的位移持續增加時,
: : 所能得到的最大sinθ值為多少?
: : 像上面那張圖就是位移持續增加,下面就是過最高點後開始減少了
: : 原本想法是用軌跡方程式算出X,y平方是距離,然後在微分大於零就是漸增了
: : 但中間數學有很大的代溝
: : 再來嘗試用向量來算
: : https://i.imgur.com/nSZ5p5G.jpeg
![](https://i.imgur.com/nSZ5p5G.jpg)
: : 還是不知道怎麼求出sin最大值,求救希望能在普物範疇內解釋
: : 答案是sinθ=2*2^0.5/3 (3分之2根號2)
: : -----
: : Sent from MeowPtt on my iPhone
: 從你的向量圖或直接從向量內積做起
: s^2 = (1/4)(g^2)t^4 + (v^2)t^2 - (vgsinθ)t^3
: d(s^2)/dt = t[(g^2)t^2 - (3vgsinθ)t + 2v^2]
: = t(t - t')(t - t"), t < t' <= t"
: 落地時間t* = (2v/g)sinθ,
: 直接比較可知t* >= t'
: 代表如果t' =/= t",則落地前必發生s從增長到縮減的轉捩點
: 這不是我們要的狀況
: 所以題意要求的狀況必然是發生在(3vgsinθ)^2 - 8(gv)^2 <= 0的條件下
: => sinθ <= (2/3)√2 = (sinθ)_max
: 這應該不是普通的普物題吧?
--
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※ 編輯: dealifeth1 (111.242.206.253 臺灣), 12/13/2023 23:45:42
※ 編輯: dealifeth1 (111.242.206.253 臺灣), 12/13/2023 23:46:47
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