2-49 할리데이 10판 솔루션 일반물리학

$$\begin{cases} x_0 = 80\ut{m}\\ x_1 = 0\ut{m}\\ v_0 = 12\ut{m/s}\\ a = -g = -9.80665\ut{m/s^2}\\ \end{cases}$$
(a)$t_{0\to1}=?$ $$\begin{aligned} \Delta x &= v_0t+\frac{1}{2}at^2,\\ \Delta x_{0\to1} &= v_0t_{0\to1}+\frac{1}{2}(-g)t_{0\to1}^2\\ -x_0 &= v_0t_1-\frac{1}{2}gt_1^2\\ \end{aligned}$$ $$\begin{aligned} t_1 &= \frac{v_0\pm\sqrt{2 g x_0+v_0^2}}{g}\\ &= \frac{12\ut{m/s}\pm\sqrt{2 (9.80665\ut{m/s^2}) 80\ut{m}+(12\ut{m/s})^2}}{9.80665\ut{m/s^2}}\\ &=\frac{800 \left(300+ \sqrt{1070665}\right)}{196133}\ut{s}(\because t_1>0)\\ &\approx 5.444180972028716\ut{s}\\ &\approx 5.4\ut{s} \end{aligned}$$
(b)$\abs{v_1}=?$ $$2a\Delta x = v^2-v_0^2,$$ $$v^2 = 2a\Delta x +v_0^2$$ $$\begin{aligned} \abs{v_1} &= \sqrt{2(-g)(-x_0) +v_0^2}\\ &= \sqrt{2gx_0+v_0^2}\\ &= \sqrt{2(9.80665\ut{m/s^2})(80\ut{m})+\(12\ut{m/s}\)^2}\\ &= \frac{1}{5}\sqrt{\frac{224258}{5}}\ut{m/s}\\ &\approx 42.35639266982022\ut{m/s}\\ &\approx 42\ut{m/s} \end{aligned}$$