2-48 할리데이 10판 솔루션 일반물리학

$$\begin{cases} x_0 = 30.0\ut{m}\\ x_1 = 0\ut{m}\\ v_0 = -15.0\ut{m/s}\\ a = -g = -9.80665\ut{m/s^2}\\ \end{cases}$$
(a)$t_{0\to1}=?$ $$\begin{aligned} \Delta x &= v_0t+\frac{1}{2}at^2,\\ \Delta x_{0\to1} &= v_0t_{0\to1}+\frac{1}{2}(-g)t_{0\to1}^2\\ -x_0 &= v_0t_1-\frac{1}{2}gt_1^2\\ \end{aligned}$$ $$\begin{aligned} t_1 &= \frac{v_0\pm\sqrt{2 g x_0+v_0^2}}{g}\\ &= \frac{-15\ut{m/s}\pm\sqrt{2 (9.80665\ut{m/s^2}) 30\ut{m}+(15\ut{m/s})^2}}{9.80665\ut{m/s^2}}\\ &=\frac{200 \left(-1500+ \sqrt{8133990}\right)}{196133}\ut{s}(\because t_1>0)\\ &\approx 1.378671694308451\ut{s}\\ &\approx 1.38\ut{s} \end{aligned}$$
(b)$v_1=?$ $$2a\Delta x = v^2-v_0^2,$$ $$v^2 = 2a\Delta x +v_0^2$$ $$\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})(30\ut{m})+\(-15\ut{m/s}\)^2}\\ &= \frac{1}{10}\sqrt{\frac{813399}{10}}\ut{m/s}\\ &\approx 28.52015077098997\ut{m/s}\\ &\approx 28.5\ut{m/s} \end{aligned}$$