2-62 할리데이 10판 솔루션 일반물리학

$$ \begin{cases} t_1 = \text{Man is 15cm From Bottom 1st Time}\\ t_2 = \text{Man is Under 15cm From Top 1st Time}\\ t_3 = \text{Man is Top Point}\\ t_4 = \text{Man is Under 15cm From Top 2st Time}\\ t_5 = \text{Man is 15cm From Bottom 2nd Time}\\ t_6 = \text{Man is Landing}\\ \end{cases} $$ $$\begin{cases} x_0=x_6=0\\ x_3=0.780\ut{m}\\ v_3=0\\ x_1=x_5=0.150\ut{m}\\ x_2=x_4=x_3-0.150\ut{m}\\ a=-g = -9.80665\ut{m/s^2}\\ \end{cases} $$ $$ 2a\Delta x = v^2-v_0^2,$$ $$ \begin{aligned} 2(-g)\Delta x_{0\to3} &= v_3^2-v_0^2\\ -2gx_3 &= -v_0^2\\ v_0^2 &=2g x_3\\ v_0 &= \sqrt{2gx_3} \ (\because v_0>0)\\ \end{aligned} $$ $$ \begin{aligned} \Delta x &= v_0t+\frac{1}{2}at^2,\\ \Delta x_{0\to t} &= v_0t+\frac{1}{2}(-g)t^2\\ x_t-x_0 &= \sqrt{2gx_3}t-\frac{1}{2}gt^2\\ x_t &= \sqrt{2gx_3}t-\frac{1}{2}gt^2\\ \end{aligned} $$ $$\therefore t = \sqrt{\frac{2}{g}}\left(\sqrt{x_3}\pm\sqrt{x_3-x_t}\right)\taag{2-62-1}$$ $$\text{(-Means UpDirection Time)}$$ $$\text{(+Means DownDirection Time)}$$
(a) $t_{2\to4}=?$ $$ \begin{aligned} \Ans &= t_{2\to4} = 2t_{2\to3}=2(t_3-t_2)\\ &= 2\bra{\sqrt{\frac{2}{g}}\left(\sqrt{x_3}-\sqrt{x_3-x_3}\right)-\sqrt{\frac{2}{g}}\left(\sqrt{x_3}-\sqrt{x_3-x_2}\right)}\\ &= 2\sqrt{\frac{2}{g}}\bra{\sqrt{x_3}-\left(\sqrt{x_3}-\sqrt{x_3-x_2}\right)}\\ &= 2\sqrt{\frac{2}{g}}\sqrt{x_3-x_2}\\ &= 2\sqrt{\frac{2}{g}}\sqrt{x_3-(x_3-0.150\ut{m})}\\ &= 2\sqrt{\frac{2}{g}}\sqrt{0.150\ut{m}}\\ &= 2\sqrt{\frac{2\cdot0.150\ut{m}}{g}}\\ &=40 \sqrt{\frac{15}{196133}}\ut{s}\\ &\approx 0.3498084412322713\ut{s}\\ &\approx 0.350\ut{s} \end{aligned} $$
(b) $t_{0\to1}+t_{5\to6}=?$ $$ \begin{aligned} \Ans &= 2t_{0\to1}=2(t_1-t_0)=2t_1\\ &= 2\sqrt{\frac{2}{g}}\left(\sqrt{x_3}-\sqrt{x_3-x_1}\right)\\ &= 2\sqrt{\frac{2}{g}}\left(\sqrt{0.780\ut{m}}-\sqrt{0.780\ut{m}-0.150\ut{m}}\right)\\ &=\sqrt{\frac{6}{(47+2 \sqrt{546})g}}\\ &=\sqrt{\frac{6}{(47+2 \sqrt{546})9.80665}}\ut{s}\\ &=40 \sqrt{\frac{3 \left(47-2\sqrt{546}\right)}{196133}}\ut{s}\\ &\approx 0.080792001405712\ut{s}\\ &\approx 0.0808\ut{s} \end{aligned} $$ 점프 상단에서의 유지시간이 훨씬 길기 때문이다.