4-38 할리데이 10판 솔루션 일반물리학

$$\begin{cases} t_0 : \text{Start}, 0\ut{sec}\\ \end{cases}$$ $$\begin{cases} v_0 &= 31\ut{m/s}\\ v_{2.5} &= 19\ut{m/s}\\ v_5 &= 31\ut{m/s}\\ \end{cases}$$ $$\begin{cases} \vec a &=-g\j\\ g &\approx 9.80665\ut{m/s^2}\\ \end{cases}$$
(a) $\Delta x_{0\rarr\text{Ground}}=?$ $$\begin{aligned} \abs{\vec v}_{\min}&=\vec v_{\text{HighMax}},\\ t_{2.5}&=t_{\text{HighMax}}\\ \end{aligned}$$ $$\begin{aligned} \vec v_{\text{HighMax}}&=(v_x,0),\\ v_{2.5} &= 19\ut{m/s},\\ \vec v_{2.5} &= (19,0)\ut{m/s}\\ v_x &= 19\ut{m/s} \end{aligned}$$ $$\begin{aligned} \abs{v_{0y}} &= \abs{v_{\text{ground}y}},\\ t_5&=t_{\text{ground}}\\ \end{aligned}$$ $$\begin{aligned} \Delta x_{0\rarr\text{Ground}} &= \Delta x_{0\rarr5}\\ &=v_x t\\ &=19 \cdot 5 \ut{m}\\ &= 95 \ut{m} \end{aligned}$$
(b) $y_{\text{HighMax}}=?$ $$\begin{cases} \vec v_0 &= (19,v_{0y})\ut{m/s}\\ \abs{\vec v_0} &= 31\ut{m/s}\\ \end{cases}$$ $$\vec v_0 = (19,10\sqrt{6})\ut{m/s} $$ $$\begin{aligned} y_{\text{HighMax}}&=\Delta y_{0\rarr2.5}\\ \Delta y&=v_{0y}t+\frac{1}{2}at^2,\\ \Delta y_{0\rarr2.5} &= \(10\sqrt{6}\) (2.5)-\frac{1}{2}g(2.5)^2\ut{m}\\ &=25 \sqrt{6}-\frac{25}{8}g\ut{m}\\ &\approx 30.5914623196\ut{m}\\ &\approx 31\ut{m} \end{aligned}$$