4-36 할리데이 10판 솔루션 일반물리학

$$\begin{cases} y_0 &= 2.42\ut{m}\\ v_x &= 23.6\ut{m/s}\\ h_N &= 0.90\ut{m}\\ \Delta x &= 12\ut{m} \end{cases}$$ $$\begin{cases} \vec a &=-g\j\\ g &\approx 9.80665\ut{m/s^2}\\ \end{cases}$$ $$\begin{cases} t_0 : \text{Start}\\ t_1 : \text{On Net}\\ \end{cases}$$
(a),(b) $\text{horizon}$ $$\Delta x = v_x t,$$ $$\therefore t = \frac{\Delta x}{v_x}$$ $$\begin{aligned} S &= v_0t+\frac{1}{2}at^2,\\ \Delta y &= v_{y0}t+\frac{1}{2}(-g)t^2\\ &=-\frac{g}{2}\(\frac{\Delta x}{v_x}\)^2\\ y&=y_0+\Delta y\\ &=y_0-\frac{g\Delta x^2}{2v_x^2}\\ &=\frac{121}{50}-\frac{450 g}{3481}\\ &\approx 1.152262999138179\ut{m}\\ &\approx 1.15\ut{m}\\ \end{aligned}$$
(a) $\text{Net Over ?}$ $$\begin{aligned} 1.15\ut{m}>0.90\ut{m},\\ \title{Ball Can Net Over} \end{aligned}$$
(b) $\text{Net Over hight?}$ $$\begin{aligned} h =&\frac{121}{50}-\frac{450 g}{3481}-0.9\\ \approx & 0.2522629991381789\ut{m}\\ \approx & 0.252\ut{m}\\ \end{aligned}$$
(c),(d) $\text{Under 5 Degree}$ $$\Delta x = v_x t,$$ $$\therefore t = \frac{\Delta x}{v_x} = \frac{\Delta x}{v_0\cos\theta}$$ $$\begin{aligned} S &= v_0t+\frac{1}{2}at^2,\\ \Delta y &= v_{y0}t+\frac{1}{2}(-g)t^2\\ &=v_{0}\(\frac{\Delta x}{v_0\cos\theta}\)\sin\theta-\frac{1}{2}g\(\frac{\Delta x}{v_0\cos\theta}\)^2\\ &=\Delta x\tan\theta-\frac{g(\Delta x)^2}{2v_0^2\cos^2\theta}\\ y&=y_0+\Delta y\\ &=y_0+\Delta x\tan\theta-\frac{g(\Delta x)^2}{2v_0^2\cos^2\theta}\\ &\approx 0.09269544029315635\ut{m}\\ &\approx 0.0927\ut{m}\\ \end{aligned}$$
(c) $\text{Under 5 degree, Net Over ?}$ $$\begin{aligned} 0.0927\ut{m}<0.90\ut{m},\\ \title{Ball Can't Net Over} \end{aligned}$$
(d) $\text{Under 5 degree, Net Over hight??}$ $$\title{Ball Can't Net Over}$$