2-31 할리데이 10판 솔루션 일반물리학

$$ \begin{cases} v_0 &= 0 \\ x_0 &= 0 \\ x_1 &= 0.500\ut{km} = 500\ut{m} \\ a_{0\to1} &= 10.0\ut{m/s^2} \\ g &=9.80665\ut{m/s^2 } \end{cases}$$ $$\max x= x_M=?$$ $$\begin{aligned} 2aS &= v^2-v_0^2, \\ 2a_{0\to1}x_{0\to1}&= v_1^2-v_0^2 \end{aligned} $$ $$ \begin{aligned} v_1 &= \sqrt{2a_{0\to1}x_{0\to1}} \ (\because v_0=0) \\ &= \sqrt{2(10.0\ut{m/s^2})(500\ut{m})} \\ &=100\ut{m/s} \end{aligned} $$ $$ \begin{aligned} \\ 2aS &= v^2-v_0^2, \\ 2(-g)x_{1toM} &= v_{x_M}^2-v_1^2, \\ x_{1toM} &= \frac{v_1^2}{2g}\ (\because v_{x_M}=0) \\ &= \frac{(100\ut{m/s})^2}{2(9.80665\ut{m/s^2})} \\ &= \frac{100000000}{196133}\ut{m} \end{aligned} $$ $$ \begin{aligned} \Ans &= x_M = x_{0\to1}+x_{1\to M} \\ &= 500\ut{m}+\frac{100000000}{196133}\ut{m} \\ &= \frac{198066500}{196133}\ut{m} \\ &\approx 1009.858106488964\ut{m} \\ &\approx 1.01\ut{km} \end{aligned} $$