2-61 할리데이 10판 솔루션 일반물리학

$$ \begin{cases} t_0 = \text{Ball is Top of Building}\\ t_1 = \text{DownBall is Top of Window}\\ t_2 = \text{DownBall is Bottom of Window}\\ t_3 = \text{Ball is Contact Road}\\ t_4 = \text{UpBall is Bottom of Window}\\ t_5 = \text{UpBall is Top of Window}\\ \end{cases} $$ $$\begin{cases} \Delta t_{1\to2}=0.125\ut{s}\\ \Delta t_{4\to5}=0.125\ut{s}\\ \Delta t_{2\to4}=2.00\ut{s}\\ x_0=h\\ x_{1\to2}=-x_{4\to5}=-1.20\ut{m}\\ v_0=0\\ v_1=-v_5\\ v_2=-v_4\\ a=-g = -9.80665\ut{m/s^2}\\ \end{cases} $$ $$ h=?$$ $$t_{2\to3}=t_{3\to4}$$ $$ (\because \text{Same Motion in 1-3 and 3-5})$$ $$t_{2\to3}=\frac{1}{2}t_{2\to4}=1.00\ut{s}$$ $$ \Delta x = v_0t+\frac{1}{2}at^2,$$ $$\begin{cases} \Delta x_{0\to1} &= v_0t_{0\to1}+\frac{1}{2}(-g)t_{0\to1}^2\\ \Delta x_{0\to2} &= v_0t_{0\to2}+\frac{1}{2}(-g)t_{0\to2}^2\\ \end{cases} $$ $$\begin{cases} x_1-x_0 &= -\frac{1}{2}g(t_1-t_0)^2\\ x_2-x_0 &= -\frac{1}{2}g(t_2-t_0)^2\\ \end{cases} $$ $$\begin{cases} x_1-h &= -\frac{1}{2}gt_1^2\\ x_2-h &= -\frac{1}{2}gt_2^2\\ \end{cases} $$ $$ \begin{aligned} x_2-x_1 &= -\frac{1}{2}g(t_2^2-t_1^2)\\ x_{1\to2} &= -\frac{1}{2}g\bra{(t_1+t_{1\to2})^2-t_1^2}\\ &= -\frac{1}{2}gt_{1\to2}(2t_1+t_{1\to2})\\ \end{aligned} $$ $$\therefore t_1=-\frac{x_{1\to2}}{t_{1\to2}g}-\frac{t_{1\to2}}{2}$$ $$ \begin{aligned} t_3&=t_1+t_{1\to2}+t_{2\to3}\\ &=-\frac{x_{1\to2}}{t_{1\to2}g}-\frac{t_{1\to2}}{2}+t_{1\to2}+t_{2\to3}\\ &=\frac{t_{1\to2}}{2}-\frac{x_{1\to2}}{t_{1\to2}g}+t_{2\to3}\\ &=\frac{0.125}{2}-\frac{-1.20}{0.125g}+1.00\\ &=\frac{48}{5 g}+\frac{17}{16} \end{aligned} $$ $$ \begin{aligned} \Delta x &= v_0t+\frac{1}{2}at^2,\\ \Delta x_{0\to3} &= v_0t_{0\to3}+\frac{1}{2}(-g)t_{0\to3}^2\\ -h &= -\frac{1}{2}gt_{0\to3}^2\\ h &= \frac{1}{2}g\(\frac{48}{5 g}+\frac{17}{16}\)^2\\ &=\frac{(85 g+768)^2}{12800 g}\\ &=\frac{(85 \cdot9.80665+768)^2}{12800 \cdot 9.80665}\\ &=\frac{41040180000121}{2008401920000}\ut{m}\\ &\approx 20.43424654768354\ut{m} \end{aligned} $$