4-14 할리데이 11판 솔루션 일반물리학

$$\begin{cases} t_1&=3.2\ut{s}\\ t_2&=t_1+2.9\ut{s}\\ x_2&=86.0\ut{m}\\ \end{cases}$$ $$\ab{a}$$ $$\begin{aligned} S&=vt-\frac{1}{2}at^2,\\ \Delta y_{0\rarr1}&=(0)t-\frac{1}{2}(-g)(3.2)^2\\ &=\frac{128}{25}g\\ &=\frac{784532}{15625}\ut{m}\\ &= 50.210048\ut{m}\\ &\approx 50\ut{m}\\ \end{aligned}$$ $$\ab{b}$$ $$\begin{aligned} S&=v_0t+\frac{1}{2}at^2,\\ \Delta y_{1\rarr2} &=(0)t+\frac{1}{2}(-g)(2.9)^2\\ &=-\frac{841}{200}g\\ \end{aligned}$$ $$\begin{aligned} \Sigma y &= \Delta y_{0\rarr1}+\Delta y_{1\rarr2}\\ &=\frac{128}{25}g-\frac{841}{200}g\\ &=\frac{183}{200}g\\ &= \frac{35892339}{4000000}\ut{m}\\ &=8.97308475\ut{m}\\ &\approx 9.0\ut{m} \end{aligned}$$ $$\ab{c}$$ $$\begin{aligned} v_x &=\frac{\Delta x_{0\rarr2}}{t_{0\rarr2}}\\ &=\frac{85}{3.2+2.9}\ut{m/s}\\ &=\frac{850}{61}\ut{m/s}\\ \end{aligned}$$ $$\begin{cases} \Delta x_{0\rarr3} &= v_xt_{0\rarr3}\\ \Delta x_{0\rarr2} &= v_xt_{0\rarr2}\\ \end{cases}$$ $$\frac{\Delta x_{0\rarr3}}{\Delta x_{0\rarr2}}= \frac{t_{0\rarr3}}{t_{0\rarr2}}$$ $$\begin{aligned} \Delta x_{0\rarr3}&=\frac{2\cdot3.2}{3.2+2.9}\cdot 86\ut{m}\\ &=\frac{5504}{61}\ut{m}\\ \end{aligned}$$ $$\begin{aligned} \Delta x_{2\rarr3}&=\Delta x_{0\rarr3}-\Delta x_{0\rarr2}\\ &=\(\frac{5504}{61}-86\)\ut{m}\\ &=\frac{258}{61}\ut{m}\\ &\approx 4.229508196721311\ut{m}\\ &\approx 4.2\ut{m}\\ \end{aligned}$$