4-3 할리데이 11판 솔루션 일반물리학

$$\begin{cases} v_{A0}=520\ut{m/s},& \theta_A=14.0\degree\\ v_{B0}=630\ut{m/s},& \theta_B=16.0\degree\\ v_{C0}=750\ut{m/s},& \theta_C=18.0\degree\\ v_{D0}=870\ut{m/s},& \theta_D=20.0\degree\\ v_{E0}=1000\ut{m/s},& \theta_E=22.0\degree\\ g=9.80665\ut{m/s^2} \end{cases}$$ $$\begin{aligned} \Delta x &= v_xt\\ &=v_0t\cos\theta \end{aligned}$$ $$\begin{aligned} \Delta y &=v_{y0}t+\frac{1}{2}(-g)t^2\\ &=v_0t\sin\theta-\frac{1}{2}gt^2\\ \end{aligned}$$ $$\ab{a}$$ $$\begin{aligned} R_{t=20}&=\(v_0t\cos\theta,v_0t\sin\theta-\frac{1}{2}gt^2\)\\ &=\(20v_0\cos\theta,20v_0\sin\theta-200g\)\\ \end{aligned}$$ $$\begin{cases} A:\(20v_{A0}\cos\theta_A,20v_{A0}\sin\theta_A-200g\)\\ B:\(20v_{B0}\cos\theta_B,20v_{B0}\sin\theta_B-200g\)\\ C:\(20v_{C0}\cos\theta_C,20v_{C0}\sin\theta_C-200g\)\\ D:\(20v_{D0}\cos\theta_D,20v_{D0}\sin\theta_D-200g\)\\ E:\(20v_{E0}\cos\theta_E,20v_{E0}\sin\theta_E-200g\)\\ \end{cases}$$ $$=\begin{cases} A:\(20\cdot520\cdot \cos14\degree,20\cdot520\cdot \sin14\degree-200g\)\\ B:\(20\cdot630\cdot \cos16\degree,20\cdot630\cdot \sin16\degree-200g\)\\ C:\(20\cdot750\cdot \cos18\degree,20\cdot750\cdot \sin18\degree-200g\)\\ D:\(20\cdot870\cdot \cos20\degree,20\cdot870\cdot \sin20\degree-200g\)\\ E:\(20\cdot1000\cdot \cos22\degree,20\cdot1000\cdot \sin22\degree-200g\)\\ \end{cases}$$ $$\approx \begin{cases} A:\(10091.07555327036, 554.6577142365445\)\ut{m}\\ B:\(12111.89736882282, 1511.70068329419\)\ut{m}\\ C:\(14265.8477444273, 2673.924915624212\)\ut{m}\\ D:\(16350.65160167481, 3989.820493866636\)\ut{m}\\ E:\(18543.67709133575, 5530.80186831824\)\ut{m}\\ \end{cases}$$ $$\approx \begin{cases} A:\(10.1, 0.555\)\ut{km}\\ B:\(12.1, 1.51\)\ut{km}\\ C:\(14.3, 2.67\)\ut{km}\\ D:\(16.4, 3.99\)\ut{km}\\ E:\(18.5, 5.53\)\ut{km}\\ \end{cases}$$ $$\ab{b}$$ $$\begin{cases} x &=v_0t\cos\theta\\ y &=v_0t\sin\theta-\frac{1}{2}gt^2\\ \end{cases}$$ $$\begin{aligned} y&=\(\frac{x}{\cos\theta}\)\sin\theta-\frac{1}{2}g\(\frac{x}{v_0\cos\theta}\)^2\\ &=(\tan\theta)x-\(\frac{g}{2{v_0}^2}\sec ^2\theta\)x^2 \end{aligned}$$