Wiki source code of 2025-Ryudai-Butsuri-Q2A(3)

Last modified by Akiyoshi Yamakawa on 2026/02/21 18:40
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 1: = 琉大2025物理大問２A問3の計算式 =
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 3: **世間に、計算途中式の資料が見当たらないので、途中式を書いてみました。(by A.YK)**
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 5: * **最初の落下までの時間をｔ,,0,,とおき、高さｈの障害物の位置まで移動するのにかかる時間をｔ、
 6: 床に衝突する直前の鉛直方向速さをV,,y　,,とおくと、反発係数e=1/2より、衝突直後の鉛直方向速さｕ=1/2V,,y,,であり、**
 7: ** 　[[image:http://www.dietpanda.com/cgi-bin/mimetex.cgi?\frac{1}{2}g{t_0}^2=8h]]　より、[[image:http://www.dietpanda.com/cgi-bin/mimetex.cgi?{t_0}=4\sqrt{\frac{h}{g}}||height="33" width="83"]]
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 9: ** 　[[image:http://www.dietpanda.com/cgi-bin/mimetex.cgi?vt=L]]　より、[[image:http://www.dietpanda.com/cgi-bin/mimetex.cgi?t=\frac{L}{v}||height="31" width="60"]]
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11: ** 　[[image:http://www.dietpanda.com/cgi-bin/mimetex.cgi?\frac{1}{2}m{v_y}^2=mg8h]]　より、[[image:http://www.dietpanda.com/cgi-bin/mimetex.cgi?u=\frac{1}{2}v_y=2\sqrt{gh}||height="26" width="130"]]
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13: * **時刻ｔでの高さｙが障害物の高さｈよりも上であればよいので、**
14: ** [[image:http://www.dietpanda.com/cgi-bin/mimetex.cgi?h<u(t-t_0)-\frac{1}{2}g(t-t_0)^2]] 　**これに、上記のｕ、ｔ、ｔ,,0,,を代入すると、**
15: \\[[image:http://www.dietpanda.com/cgi-bin/mimetex.cgi?h<2\sqrt{gh}(\frac{L}{v}-4\sqrt{hg})-\frac{1}{2}g(\frac{L}{v}-4\sqrt{hg})^2||height="30" width="339"]]　[[image:http://www.dietpanda.com/cgi-bin/mimetex.cgi?h<2\sqrt{gh}||alt="http://www.dietpanda.com/cgi-bin/mimetex.cgi?h<2\sqrt{gh}(\frac{L}{v}-4\sqrt{hg})-\frac{1}{2}g(\frac{L}{v}-4\sqrt{hg})^2" height="20" width="76"]][[image:http://www.dietpanda.com/cgi-bin/mimetex.cgi?(\frac{L}{v}-4\sqrt{hg})-\frac{1}{2}g||alt="http://www.dietpanda.com/cgi-bin/mimetex.cgi?h<2\sqrt{gh}(\frac{L}{v}-4\sqrt{hg})-\frac{1}{2}g(\frac{L}{v}-4\sqrt{hg})^2" height="30" width="141"]][[image:http://www.dietpanda.com/cgi-bin/mimetex.cgi?(\frac{L}{v}-4\sqrt{hg})^2||alt="http://www.dietpanda.com/cgi-bin/mimetex.cgi?h<2\sqrt{gh}(\frac{L}{v}-4\sqrt{hg})-\frac{1}{2}g(\frac{L}{v}-4\sqrt{hg})^2" height="30" width="137"]]
16: \\[[image:http://www.dietpanda.com/cgi-bin/mimetex.cgi?h<2\frac{L}{v}\sqrt{gh}-8h-\frac{1}{2}g\frac{L^2}{v^2}-8h+4\frac{L}{v}\sqrt{gh}||height="40" width="345"]]
17: \\[[image:http://www.dietpanda.com/cgi-bin/mimetex.cgi?0<6\sqrt{gh}\frac{L}{v}-17h-\frac{1}{2}g\frac{L^2}{v^2}||height="43" width="250"]]　**これを、ｖの二次不等式に直すと、**
18: \\[[image:http://www.dietpanda.com/cgi-bin/mimetex.cgi?0>34v^2-12\sqrt{\frac{g}{h}}Lv+\frac{g}{h}L^2||alt="http://www.dietpanda.com/cgi-bin/mimetex.cgi?0>34v^2-12\sqrt{gh}v+\frac{g}{h}" height="40" width="200"]] 　**さらに、＝０の二次方程式としてvを求めると、**
19: \\[[image:http://www.dietpanda.com/cgi-bin/mimetex.cgi?v=\frac{6\frac{g}{h}L\pm\sqrt{36\frac{g}{h}L^2-34\frac{g}{h}L^2}}{34}=\frac{(6\pm\sqrt{2})\sqrt{\frac{g}{h}}L}{34}||alt="http://www.dietpanda.com/cgi-bin/mimetex.cgi?v=\frac{6\frac{g}{h}\pm\sqrt{36\frac{g}{h}-34\frac{g}{h}}}{34}=\frac{(6\pm\sqrt{2})\sqrt{\frac{g}{h}}}{34}" height="46" width="320"]] 　**ｖの下限なので、－の方をとって、**
20: \\[[image:http://www.dietpanda.com/cgi-bin/mimetex.cgi?v=\frac{(6-sqrt{2})}{34}\sqrt{\frac{g}{h}}L=\frac{(6-sqrt{2})(6+sqrt{2})}{34(6+sqrt{2})}\sqrt{\frac{g}{h}}L=\frac{34}{34(6+\sqrt{2})}\sqrt{\frac{g}{h}}L=\frac{1}{(6+\sqrt{2})}\sqrt{\frac{g}{h}}L||alt="http://www.dietpanda.com/cgi-bin/mimetex.cgi?v=\frac{(6-sqrt{2})}{34}\sqrt{\frac{g}{h}}}=\frac{(6-sqrt{2})(6+sqrt{2})}{34(6+sqrt{2})}}\sqrt{\frac{g}{h}}=\frac{34}{34(6+\sqrt{2})}\sqrt{\frac{g}{h}}=\frac{1}{(6+\sqrt{2})}\sqrt{\frac{g}{h}}" height="50" width="500"]]　　**・・・答え**
21: \\**（注：次元解析して速度の単位になっていることも確認！　√{(m/s^^2^^)/m)}m=m/s）**