2-44 할리데이 10판 솔루션 일반물리학

$$\begin{cases} t_2 = 0.200\ut{s}\\ x_2 = 0.588\ut{m}\\ x_1 = 0.544\ut{m}\\ a = -g = -9.80665\ut{m/s^2}\\ \end{cases}$$
(a) $\abs{v_0}=?$ $$\Delta x = v_0t+\frac{1}{2}at^2,$$ $$\begin{aligned} \abs{v_0}&= \frac{x_2-\frac{1}{2}(-g)t_2^2}{t_2}\\ &= \frac{x_2}{t_2}+\frac{gt_2}{2}\\ &= \frac{0.588\ut{m}}{0.200\ut{s}}+\frac{9.80665\ut{m/s^2}(0.200\ut{s})}{2}\\ &= \frac{254933}{20000}\ut{m/s}\\ &= 12.74665\ut{m/s}\\ &\approx 12.7\ut{m/s} \end{aligned}$$
(b)$v_1=?$ $$2a\Delta x = v^2-v_0^2,$$ $$v^2 =2a\Delta x+v_0^2$$ $$v =\pm\sqrt{2a\Delta x+v_0^2}$$ $$\begin{aligned} v_1 &=\pm\sqrt{2(-g)(\Delta x_{0\to1})+v_0^2}\\ &=\pm\sqrt{v_0^2-2gx_1}\\ &=\pm\sqrt{\( \frac{254933}{20000}\ut{m/s}\)^2-2(9.80665\ut{m/s^2})(0.544\ut{m})}\\ &=\pm \frac{\sqrt{60722980409}}{20000}\ut{m/s}\\ &\approx \pm12.32101663916172\ut{m/s}\\ &\approx \pm12.3\ut{m/s}\\ \end{aligned}$$
(c)$\max \Delta x=?$ $$ t_3 := t_{\max x},$$ $$ v_3=0 $$ $$2a\Delta x = v^2-v_0^2,$$ $$\begin{aligned} x_{\max}&=x_3 = \frac{v_3^2-v_0^2}{2(-g)}\\ &= \frac{v_0^2-v_3^2}{2g}\\ &= \frac{\(\frac{\sqrt{60722980409}}{20000}\ut{m/s}\)^2-0^2}{2\(9.80665\ut{m/s^2}\)}\\ &=\frac{8674711487}{1120760000}\ut{m}\\ &\approx 7.740025952924801\ut{m}\\ &\approx 7.74\ut{m}\\ \end{aligned}$$