4-39 할리데이 10판 솔루션 일반물리학

$$\begin{cases} d_{1.50} &= -25.0\ut{m}\\ \theta_{1.50} &= 60.0\degree\\ \end{cases}$$ $$\begin{cases} \vec a &=-g\j\\ g &\approx 9.80665\ut{m/s^2}\\ \end{cases}$$ (a) $h_0=?$ $$\begin{aligned} d&=v_x t\\ v_x&=\frac{25}{1.5}\\ &= \frac{50}{3} \end{aligned}$$ $$\begin{aligned} \vec v_{1.5}&=(-v_{x1.5},-v_{y1.5})\\ &=(-v_x,-\sqrt{3}v_x)\\ &=\(-\frac{50}{3},-\frac{50}{\sqrt{3}}\)\\ \end{aligned}$$ $$\begin{aligned} \Delta y&=v_{y}t-\frac{1}{2}at^2,\\ -h_0&=\(-\frac{50}{\sqrt{3}}\)(1.5)-\frac{1}{2}(-g)(1.5)^2\\ \end{aligned}$$ $$\begin{aligned} h_0&=25 \sqrt{3}-\frac{9}{8}g\\ &\approx 32.2687889392\ut{m}\\ &\approx 32.3\ut{m} \end{aligned}$$ (b) $\abs{\vec v_0}=?$ $$\begin{aligned} v&=v_0+at,\\ v_y&=v_{0y}+a_yt\\ -\frac{50}{\sqrt{3}}&=v_{0y}+(-g)(1.5)\\ v_{0y}&=\frac{3 g}{2}-\frac{50}{\sqrt{3}}\\ \end{aligned}$$ $$\vec v_0=\(-\frac{50}{3},\frac{3 g}{2}-\frac{50}{\sqrt{3}}\) $$ $$\begin{aligned} \abs{\vec v_0}&=\sqrt{\(-\frac{50}{3}\)^2+\(\frac{3 g}{2}-\frac{50}{\sqrt{3}}\)^2}\\ &\approx 21.8680971511\ut{m/s}\\ &\approx 21.9\ut{m/s} \end{aligned}$$ (c) $\abs{\theta_0}=?$ $$\vec v_0=\(-\frac{50}{3},\frac{3 g}{2}-\frac{50}{\sqrt{3}}\)$$ $$\begin{aligned} \abs{\theta_0} &=\abs{\tan^{-1}\(\frac{\frac{3 g}{2}-\frac{50}{\sqrt{3}}}{-\frac{50}{3}}\)}\\ &=\abs{\tan^{-1}\(\frac{-100\sqrt3+9g}{-100}\)}\\ &\approx \abs{-2.4374166389713903}\ut{rad}\\ &\approx 2.44\ut{rad} \end{aligned}$$ (d) $\theta_0>0?$ $$\title{Down}$$