4-24 할리데이 10판 솔루션 일반물리학

$$\begin{cases} \vec a &= -g\j\\ g &= 9.8\ut{m/s^2}\\ \end{cases}$$ $$\begin{cases} \Delta x &= 8.95\ut{m}\\ v_0 &= 9.5\ut{m/s}\\ \end{cases}$$ $$\begin{cases} v_x &= v_0\cos\theta\\ v_y &= v_0\sin\theta\\ \end{cases}$$ $$\begin{aligned} \Delta y &= v_{y0}t+\frac{1}{2}a_yt^2,\\ 0 &= (v_0\sin\theta) t+\frac{1}{2}(-g)t^2\\ 0 &= v_0\sin\theta -\frac{1}{2}gt\\ t &= \frac{2v_0}{g}\sin\theta\\ \end{aligned}$$ $$\begin{aligned} \Delta x &= v_xt,\\ &= (v_0\cos\theta)\(\frac{2v_0}{g}\sin\theta\)\\ &= \frac{v_0^2}{g}\sin(2\theta)\\ \end{aligned}$$ $$\begin{aligned} &\text{at max }\Delta x,\\ 2\theta&=\frac{\pi}{2}\ut{rad}\\ \therefore \theta&=\frac{\pi}{4}\ut{rad}\\ \end{aligned}$$ $$\title{Find max x}$$ $$\begin{aligned} \Delta y &= v_{y0}t+\frac{1}{2}a_yt^2,\\ 0 &= \(v_0\sin\frac{\pi}{4}\)t+\frac{1}{2}(-g)t^2\\ 0 &= \frac{v_0}{\sqrt{2}}-\frac{1}{2}gt\\ t &=\frac{v_0\sqrt{2}}{g}\\ \end{aligned}$$ $$\begin{aligned} \max \Delta x &= v_xt,\\ &= \(v_0\cos\frac{\pi}{4}\)\(\frac{v_0\sqrt{2}}{g}\)\\ &= \frac{v_0^2}{g}\\ \end{aligned}$$ $$\begin{aligned} \Ans &= \max \Delta x-\Delta x\\ &=\frac{(9.5\ut{m/s})^2}{9.8\ut{m/s^2}}-8.95\ut{m}\\ &=\frac{127}{490}\ut{m}\\ &\approx 0.259183673469\ut{m}\\ &\approx 0.26\ut{m}\\ \end{aligned}$$