2-11 할리데이 11판 솔루션 일반물리학

$$\begin{cases} H&=S_{1.70}\\ h_{M}&=S_{1.70+1.40}=S_{3.10}\\ a&=-g=-9.80665\ut{m/s^2} \end{cases}$$ $$v=v_0+at,$$ $$\begin{aligned} v_0&=v_{3.1}-at\\ &=(0)-(-g)(3.1)\\ &=3.1g \end{aligned}$$ $$S=v_0t+\frac{1}{2}at^2,$$ $$\begin{aligned} H&=(3.1g)(1.7)+\frac{1}{2}(-g)(1.7)^2\\ &=\frac{153}{40}g\\ &=\frac{153}{40}\cdot9.80665\ut{m}\\ &=\frac{30008349}{800000}\ut{m}\\ &=37.51043625\ut{m}\\ &\approx37.5\ut{m} \end{aligned}$$