2-14 할리데이 11판 솔루션 일반물리학

$$\begin{cases} t_1 &= 1.12\ut{s} &=\text{Time at }h_{\max}\\ t_2 &= 6.00\ut{s} &=\text{Time at Ground}\\ a&=-g&=-9.80665\ut{m/s^2} \end{cases}$$ $$\ab{a}$$ $$v=v_0+at,$$ $$\begin{aligned} v_0&=v-at\\ &=0-(-g)t_1\\ &=9.80665\ut{m/s^2}\cdot1.12\ut{s}\\ &=\frac{1372931}{125000}\ut{m/s}\\ &=10.983448\ut{m/s}\\ &\approx11.0\ut{m/s} \end{aligned}$$ $$\ab{b}$$ $$S=vt-\frac{1}{2}at^2,$$ $$\begin{aligned} h_{\max}&=(0)t-\frac{1}{2}(-g)(1.12\ut{s})^2\\ &=\frac{9610517}{1562500}\ut{m}\\ &=6.15073088\ut{m}\\ &\approx6.15\ut{m} \end{aligned}$$ $$\ab{c}$$ $$S=v_0t+\frac{1}{2}at^2,$$ $$\begin{aligned} -H&=v_0t+\frac{1}{2}at^2\\ &=(10.983448\ut{m/s})(6.00\ut{s})+\frac{1}{2}(-g)(6.00\ut{s})^2\\ &=-\frac{27654753}{250000}\ut{m}\\ &=-110.619012\ut{m}\\ &\approx -111\ut{m}\\ \end{aligned}$$ $$\Ans = H\approx 111\ut{m}$$