4-42 할리데이 10판 솔루션 일반물리학

$$\begin{cases} \text{StartPoint}=O&=(0,0)\ut{m}\\ \text{1stTower}=A&=(23,15)\ut{m}\\ \text{2ndTower}=B&=(23+\frac{23}{2},15)\ut{m}\\ \text{Goal}=C&=(R,0)\ut{m}\\ \end{cases}$$ $$\begin{cases} v_0 &= 26.5\ut{m/s}\\ \theta_0 &= 53\degree\\ \end{cases}$$ $$\begin{cases} t_0&=\text{Start Time}\\ t_1&=\text{Over 1stTower Time}\\ t_2&=\text{Over 2ndTower Time}\\ t_3&=\text{Goal Time}\\ \end{cases}$$ $$\begin{cases} \vec a &=-g\j\\ g &\approx 9.80665\ut{m/s^2}\\ \end{cases}$$
(a) $H_{1A}=?$ $$\begin{aligned} \Delta x &= v_x t\\ &= v_0t\cos\theta_0\\ t&=\frac{\Delta x}{v_0\cos\theta_0} \end{aligned}$$ $$\begin{aligned} \Delta y&=v_{0y}t+\frac{1}{2}at^2\\ H_1&=(v_0\sin\theta_0)\(\frac{\Delta x}{v_0\cos\theta_0}\)+\frac{1}{2}(-g)\(\frac{\Delta x}{v_0\cos\theta_0}\)^2\\ &=\Delta x \tan \theta_0 -\frac{g}{2}\(\frac{\Delta x }{ v_0\cos\theta_0}\)^2\\ \end{aligned}$$ $$\begin{aligned} \Ans &= H_1 - A_y\\ &=\Delta x \tan \theta_0 -\frac{g}{2}\(\frac{\Delta x }{ v_0\cos\theta_0}\)^2 - A_y\\ &=23 \tan \theta_0 +\frac{1058 }{2809}g \sec ^2\theta_0-15\\ &\approx 25.720349003\ut{m}\\ &\approx 26\ut{m} \end{aligned}$$
(b) $H_2=\max, H_{2B}=?$ $$\begin{aligned} 2a\Delta y&=v^2-v_0^2\\ \Ans &= \Delta y-B_y\\ &=\frac{v^2-v_0^2}{2a}-B_y\\ &=\frac{-(v_0\sin\theta)^2}{2(-g)}-B_y\\ &=\frac{2809 \sin ^2\theta}{8 g}-15\\ &\approx 7.83696096839\ut{m}\\ &\approx 7.8\ut{m}\\ \end{aligned}$$
(c) $R=?$ $$\begin{aligned} \Delta y&=v_{0y}t+\frac{1}{2}at^2\\ 0&=v_{0y}+\frac{1}{2}(-g)t\\ t&=\frac{2v_{0y}}{g} \end{aligned}$$ $$\begin{aligned} \Delta x &= v_x t\\ &= (v_0\cos\theta_0)\(\frac{2v_0\sin\theta}{g}\)\\ &=\frac{v_0^2 }{g}\sin (2 \theta )\\ &=\frac{2809 }{4 g}\sin (2 \theta )\\ &\approx 68.8355377191\ut{m}\\ &\approx 69\ut{m}\\ \end{aligned}$$