4-54 할리데이 11판 솔루션 일반물리학

$$\begin{cases} v_0&=82\ut{m/s}, \theta_0=45\degree\\ \Delta x_1 &= 686\ut{m}\\ h&=40.0\ut{m} \end{cases}$$ $$\begin{aligned} S&=v_0t+\frac{1}{2}at^2,\\ 0&=v_{y0}t+\frac{1}{2}(-g)t^2\\ &=v_0\sin\theta-\frac{1}{2}gt\\ t&=\frac{2v_0}{g}\sin\theta \end{aligned}$$ $$\begin{aligned} \Delta x_1 &= v_x t\\ &= v_0\cos\theta \(\frac{2v_0}{g}\sin\theta\)\\ &=\frac{{v_0}^2}{g}\sin (2\theta )\\ &=\frac{{v_0}^2}{g} \end{aligned}$$ $$\begin{aligned} g&=\frac{{v_0}^2}{\Delta x_1}\\ &=\frac{{82}^2}{686}\\ &=\frac{3362}{343}\ut{m/s^2} \end{aligned}$$ $$\begin{aligned} S&=v_0t+\frac{1}{2}at^2,\\ -h&=(v_0\sin\theta)t+\frac{1}{2}(-g)t^2\\ -h&=\frac{v_0t}{\sqrt2}-\frac{1}{2}gt^2\\ \end{aligned}$$ $$\begin{aligned} t&=\frac{\sqrt{4 gh+{v_0}^2}+v_0}{\sqrt{2} g} \end{aligned}$$ $$\begin{aligned} \Delta x_2 &= v_x t\\ &=\frac{v_0}{\sqrt2}\(\frac{\sqrt{4 gh+{v_0}^2}+v_0}{\sqrt{2} g}\)\\ &=\frac{v_0 }{2 g}\(\sqrt{4 g h+{v_0}^2}+v_0\)\\ \end{aligned}$$ $$\begin{aligned} \Ans&=\Delta x_2-\Delta x_1\\ &=\frac{v_0 }{2 g}\(\sqrt{4 g h+{v_0}^2}+v_0\)-\frac{{v_0}^2}{g}\\ &=\frac{v_0 }{2 g}\(\sqrt{4 g h+{v_0}^2}-v_0\)\\ &=21 \sqrt{329}-343\ut{m}\\ &\approx 37.90550009155811\ut{m}\\ &\approx 38\ut{m}\\ \end{aligned}$$