4-32 할리데이 10판 솔루션 일반물리학

(풀이자주 : 속력이 m단위라 임의로 m/s로 간주하여 풀었습니다.) $$\begin{cases} \theta_0 &= 30\degree\\ v_0 &= 25.0\ut{m/s}\\ d &= 20.0\ut{m}\\ \end{cases}$$ $$\begin{cases} \vec a &=-g\j\\ g &\approx 9.80665\ut{m/s^2}\\ \end{cases}$$ $$\begin{aligned} \Delta t_{0\rarr1} &= \frac{\Delta x_{0\rarr1}}{v_x}\\ &=\frac{d}{v_0\cos\theta}\\ &=\frac{8}{5 \sqrt{3}} \end{aligned}$$
(a) $h_1=?$ $$\Delta r = v_{0}t+\frac{1}{2}at^2,$$ $$\begin{aligned} \Ans &= h_1 = \Delta y_{0\rarr1}\\ &= v_{y0}t+\frac{1}{2}a_yt^2\\ &= v_0\sin\theta \(\frac{d}{v_0\cos\theta}\)+\frac{1}{2}(-g)\(\frac{d}{v_0\cos\theta}\)^2\\ &=d \tan\theta-\frac{d^2 g }{2 v_0^2\cos ^2\theta}\\ &=\frac{20}{\sqrt{3}}-\frac{32 g}{75}\\ &\approx 7.36283471712585\ut{m}\\ &\approx 7.36\ut{m} \end{aligned}$$
(b) $v_{x1}=?$ $$\begin{aligned} v_{x1} &= v_x = v_0\cos\theta\\ &=\frac{25 \sqrt{3}}{2}\ut{m/s}\\ &\approx 21.6506350946\ut{m/s}\\ &\approx 21.7\ut{m/s} \end{aligned}$$
(c) $v_{y1}=?$ $$v = v_{0}+at,$$ $$\begin{aligned} v_y&= v_{y0}+a_yt\\ &=v_0\sin\theta+(-g)\(\frac{d}{v_0\cos\theta}\)\\ &=\frac{25}{2}-\frac{8 g}{5 \sqrt{3}}\\ &\approx 3.44100477224\ut{m/s}\\ &\approx 3.44\ut{m/s}\\ \end{aligned}$$
(d) $t_{0\rarr1}>t_{\max} ?$ $$v = v_{0}+at,$$ $$\begin{aligned} v_y &= v_{y0}+a_yt\\ 0 &= v_0\sin\theta +(-g)t_{\max}\\ \end{aligned}$$ $$\begin{aligned} t_{\max} &= \frac{v_0\sin\theta}{g}\\ &=\frac{25}{2 g}\\ \end{aligned}$$ $$\begin{aligned} t_{0\rarr1}&>t_{\max}\\ \frac{8}{5 \sqrt{3}}&>\frac{25}{2 g}\\ 0.923760430703 &> 1.27464526622\\ \end{aligned}$$ $$\title{Not vertical MaxTime}$$