2-57 할리데이 10판 솔루션 일반물리학

$$ \begin{cases} t_1 = \text{Contact Start Floor}\\ t_2 = \text{Contact End Floor}\\ t_3 = \text{ReUprising HighPoint}\\ \end{cases} $$ $$ \begin{aligned} t_{1\to2} &= 12.0\ut{ms}\\ &= 12.0\ut{ms}\cdot\frac{1\ut{s}}{1000\ut{ms}}\\ &= \frac{3}{250}\ut{s}\\ \end{aligned} $$ $$\begin{cases} x_0 = 4.00\ut{m}\\ x_3 = 2.00\ut{m}\\ x_1 = x_2 = 0\\ v_0 = v_3 = 0\\ t_{1\to2} = \frac{3}{250}\ut{s}\\ a=-g = -9.80665\ut{m/s^2}\\ \end{cases} $$
(a) $\bar a_{1\to2}=?$ $$ 2a\Delta x = v^2-v_0^2,$$ $$ 2(-g)\Delta x_{0\to1} = v_1^2-v_0^2,$$ $$ \begin{aligned} v_1^2 &= 2(-g)(x_1-x_0) + v_0^2\\ v_1 &= \pm\sqrt{2(-g)(0-x_0) + 0^2}\\ \end{aligned} $$ $$\therefore v_1 = -\sqrt{2gx_0}(\because v_1<0)$$ $$ 2a\Delta x = v^2-v_0^2,$$ $$ 2(-g)\Delta x_{2\to3} = v_3^2-v_2^2,$$ $$ \begin{aligned} v_2^2 &= v_3^2-2(-g)(x_3-x_2)\\ v_2 = &\pm\sqrt{0^2-2(-g)(x_3-0)}\\ \end{aligned} $$ $$\therefore v_2 = \sqrt{2gx_3}(\because v_2>0)$$ $$ \begin{aligned} \bar a_{1\to2} &= \frac{v_2-v_1}{\Delta t_{1\to2}}\\ &= \frac{\sqrt{2gx_3}-\(-\sqrt{2gx_0}\)}{\Delta t_{1\to2}}\\ &= \frac{\sqrt{2g}\(\sqrt{x_3}+\sqrt{x_0}\)}{\Delta t_{1\to2}}\\ &= \frac{\sqrt{2(9.80665\ut{m/s^2})}\(\sqrt{2.00\ut{m}}+\sqrt{4.00\ut{m}}\)}{\frac{3}{250}\ut{s}}\\ &=\frac{5}{6} \sqrt{196133} \left(2+\sqrt{2}\right)\ut{m/s^2}\\ &\approx 1260.041278676706\ut{m/s^2}\\ &\approx 1.26\ut{km/s^2} \end{aligned} $$
(b) $\text{Up Direction}$