2-54 할리데이 10판 솔루션 일반물리학

$$\begin{cases} A = \text{1st Rock}\\ B = \text{2st Rock}\\ t_0 = \text{1st Rock Start}\\ t_1 = \text{2nd Rock Start}\\ t_2 = \text{A,B Rocks Landing}\\ \end{cases}$$ $$\begin{cases} x_{A0} = x_{B0}=x_{B1}=x_0=53.6\ut{m}\\ x_{A2} = x_{B2}=x_2=0\ut{m}\\ t_1 = 1.00\ut{s}\\ v_{A0} = 0\\ a=a_A = a_B =-g = -9.80665\ut{m/s^2}\\ \end{cases} $$
(a)$v_{B1}=?$ $$ \Delta x = v_0t+\frac{1}{2}at^2,$$ $$ \begin{aligned} \Delta x_{A0\to2} &= \frac{1}{2}(-g)t_{0\to2}^2\\ x_{A2}-x_{A0} &= \frac{1}{2}(-g)t_2^2\\ x_0 &= \frac{1}{2}gt_2^2\\ \end{aligned} $$ $$\therefore t_2=\sqrt{\frac{2x_0}{g}}$$ $$ \begin{aligned} t_{1\to2}&=t_2-t_1\\ &=\sqrt{\frac{2x_0}{g}}-t_1\\ &=\sqrt{\frac{2(53.6\ut{m})}{9.80665\ut{m/s^2}}}-1.00\ut{s}\\ &=80 \(\sqrt{\frac{335}{196133}}-1\)\ut{s} \end{aligned} $$ $$ \Delta x = v_0t+\frac{1}{2}at^2,$$ $$ \begin{aligned} \Delta x_{B1\to2} &= v_{B1}t_{1\to2}+ \frac{1}{2}(-g)t_{1\to2}^2\\ x_{B2}-x_{B1} &= v_{B1}t_{1\to2}-\frac{1}{2}gt_{1\to2}^2\\ \frac{1}{2}gt_{1\to2}^2-x_{B1} &= v_{B1}t_{1\to2}\\ \end{aligned} $$ $$ \begin{aligned} v_{B1}&=\frac{1}{2}gt_{1\to2}-\frac{x_{B1}}{t_{1\to2}}\\ &=\frac{1}{2}(9.80665\ut{m/s^2})\(80 \sqrt{\frac{335}{196133}}-1\)\ut{s}-\frac{53.6\ut{m}}{\(80 \sqrt{\frac{335}{196133}}-1\)\ut{s}}\\ &=-\frac{196133 \left(4091867+80\sqrt{65704555}\right)}{77914680000}\ut{m/s}\\ &\approx -11.93274339755394\ut{m/s}\\ &\approx -11.9\ut{m/s} \end{aligned} $$ $$\therefore \abs{v_{B1}}\approx 11.9\ut{m/s}$$
(b) $$ v=v_0+at,$$ $$ \begin{aligned} v_A(t)&=v_{A0}+a_At\\ &=-gt \end{aligned} $$ $$ \begin{aligned} v_{B|t>1}(t) &= v_{B1}-g(t-t_1)\\ &= v_{B1}+gt_1-gt\\ &=-\frac{196133 \left(4091867+80\sqrt{65704555}\right)}{77914680000}+g-gt\\ &=-\frac{196133 \left(196133+80\sqrt{65704555}\right)}{77914680000}-gt \end{aligned} $$ $$ v_B(t)= \begin{cases} 0, &(t<1.00)\\ v_{B|t>1}(t), &(t>1.00)\\ \end{cases} $$