2-55 할리데이 10판 솔루션 일반물리학

$$\begin{cases} t_0 = \text{Rock Fall Start}\\ t_1 = \text{Rock Contact Start}\\ t_2 = \text{Rock Stop}\\ \end{cases}$$ $$\begin{aligned} t_{1\to2} &= 20.0\ut{ms}\\ &=20.0\ut{ms}\cdot\frac{1\ut{s}}{1000\ut{ms}}\\ &=\frac{1}{50}\ut{s} \\ \end{aligned}$$ $$\begin{cases} x_0 = 15.0\ut{m}\\ x_1 = x_2=0\ut{m}\\ t_{1\to2} = \frac{1}{50}\ut{s}\\ v_0 = 0\\ v_2 = 0\\ a=-g = -9.80665\ut{m/s^2}\\ \end{cases}$$
(a)$\bar a =?$ $$2a\Delta x = v^2-v_0^2,$$ $$2(-g)(x_1-x_0) = v_1^2-v_0^2$$ $$\begin{aligned} v_1^2&=-2g(-x_0)\\ v_1&=-\sqrt{2gx_0}(\because v_1<0)\\ \end{aligned}$$ $$\begin{aligned} \bar a &= \frac{\Delta v_{1\to2}}{\Delta t_{1\to2}}\\ &= \frac{v_2-v_1}{\Delta t_{1\to2}}\\ &= \frac{-v_1}{\Delta t_{1\to2}}\\ &= \frac{\sqrt{2gx_0}}{\Delta t_{1\to2}}\\ &= \frac{\sqrt{2(9.80665\ut{m/s^2})(15.0\ut{m})}}{\frac{1}{50}\ut{s}}\\ &= \frac{\sqrt{2941995}}{2}\ut{m/s^2}\\ &\approx 857.6122375526132\ut{m/s^2}\\ &\approx 858\ut{m/s^2}\\ \end{aligned}$$
(b) $$\text{Up Direction}$$