2-58 할리데이 10판 솔루션 일반물리학

$$\begin{cases} t_1 = \text{Landing Before 1 second}\\ t_2 = \text{Landing}\\ \end{cases}$$ $$\begin{cases} t_{1\to2} = 1\ut{s}\\ x_0 = h\\ x_1 = 0.6h\\ x_2 = 0\\ v_0 = 0\\ a=-g = -9.80665\ut{m/s^2}\\ \end{cases}$$ $$t_2-t_1 = 1\ut{s},$$ $$\therefore t_1 = t_2-1\ut{s}$$ $$\Delta x = v_0t+\frac{1}{2}at^2,$$ $$\title{Make System of Equations}$$ $$\begin{cases} \Delta x_{0\to1} &= v_0t_{0\to1}+\frac{1}{2}at_{0\to1}^2\\ \Delta x_{0\to2} &= v_0t_{0\to2}+\frac{1}{2}at_{0\to2}^2\\ \end{cases}$$ $$\begin{cases} x_1-x_0 &= \frac{1}{2}(-g)t_1^2\\ x_2-x_0 &= \frac{1}{2}(-g)t_2^2\\ \end{cases}$$ $$\begin{cases} 0.6h-h &= -\frac{1}{2}g(t_2-1\ut{s})^2\\ -h &= -\frac{1}{2}gt_2^2\\ \end{cases}$$ $$\begin{cases} 0.4\underset{x}{h} &= \frac{1}{2}g(\underset{y}{t_2}-1\ut{s})^2\\ \underset{x}{h} &= \frac{1}{2}g\underset{y}{t_2}^2\\ \end{cases}$$ $$\text{(Underset x,y Means Unknown Value)}$$ $$t_2=\frac{1}{3} \left(5\pm\sqrt{10}\right)$$ $$\begin{aligned} h&=\frac{5}{18} \left(7\pm2 \sqrt{10}\right) g\\ &=\frac{5}{18} \left(7\pm \sqrt{10}\right) (9.80665)\ut{m}\\ &=\frac{196133 \left(7\pm2 \sqrt{10}\right)}{72000}\ut{m}\\ \end{aligned}$$
(a) $t_{0\to2}=?$ $$\begin{aligned} t_{2+}&=\frac{1}{3} \left(5+\sqrt{10}\right)\ut{s}\\ &\approx 2.720759220056126\ut{s}\\ &\approx 3\ut{s}\\ t_{2-}&=\frac{1}{3} \left(5-\sqrt{10}\right)\ut{s}\\ &\approx 0.6125741132772068\ut{s}\\ &\approx 1\ut{s}\\ \end{aligned}$$ $$\therefore t_2\approx 3\ut{s}$$
(b) $h=?$ $$\begin{aligned} h_+&=\frac{196133 \left(7+2 \sqrt{10}\right)}{72000}\ut{m}\\ &\approx 36.29701400893902\ut{m}\\ &\approx 40\ut{m}\\ h_-&=\frac{196133 \left(7-2 \sqrt{10}\right)}{72000}\ut{m}\\ &\approx 1.8399582132832\ut{m}\\ &\approx 2\ut{m}\\ \end{aligned}$$ $$\therefore h\approx 40\ut{m}$$
(c) 마지막 1초동안 낙하해야 하므로 전체 낙하시간은 1초를 초과해야 한다.