2-47 할리데이 10판 솔루션 일반물리학

$$ \begin{cases} x_0 = 98\ut{m}\\ x_1 = 0\ut{m}\\ v_0 = 14\ut{m/s}\\ a = -g = -9.80665\ut{m/s^2}\\ \end{cases} $$
(a)$\abs{v_1}=?$ $$ 2a\Delta x = v^2-v_0^2, $$ $$ v^2 = 2a\Delta x +v_0^2 $$ $$ \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})(98\ut{m})+\(14\ut{m/s}\)^2}\\ &= \frac{7 }{50}\sqrt{\frac{216133}{2}}\ut{m/s}\\ &\approx 46.02285736457484\ut{m/s}\\ &\approx 46\ut{m/s} \end{aligned} $$
(b)$t_{0\to1}=?$ $$ \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{14\ut{m/s}\pm\sqrt{2 (9.80665\ut{m/s^2}) 98\ut{m}+(14\ut{m/s})^2}}{9.80665\ut{m/s^2}}\\ &=\frac{200 \left(200+\sqrt{432266}\right)}{28019}\ut{s}(\because t_1>0)\\ &\approx 6.120628080391859\ut{s}\\ &\approx 6.1\ut{s} \end{aligned} $$