4-29 할리데이 11판 솔루션 일반물리학

$$\begin{cases} t&=4.70\ut{s}\\ \Delta x &= 46\ut{m}\\ y_0&=150\ut{cm}\\ g&=9.80665\ut{m/s^2} \end{cases}$$ $$\begin{aligned} v_x &= \frac{\Delta x}{t}\\ &= \frac{46}{4.70}\ut{m/s}\\ &= \frac{460}{47}\ut{m/s}\\ \end{aligned}$$ $$\begin{aligned} S&=v_0t+\frac{1}{2}at^2,\\ \Delta y&=v_0t+\frac{1}{2}(-g)t^2\\ -1.5\ut{m}&=v_{y0}(4.7)+\frac{1}{2}(-g)(4.7)^2\\ v_{y0}&=\frac{47 g}{20}-\frac{15}{47} \end{aligned}$$ $$\therefore \vec v_0= \frac{460}{47}\ut{m/s}\i+\(\frac{47 g}{20}-\frac{15}{47}\)\j$$ $$\ab{a}$$ $$\begin{aligned} v_0&=\sqrt{\(\frac{460}{47}\)^2+\(\frac{47 g}{20}-\frac{15}{47}\)^2}\\ &\approx 24.74434840758445\ut{m/s}\\ &\approx 24.7\ut{m/s}\\ \end{aligned}$$ $$\ab{b}$$ $$\begin{aligned} \Ans&=\tan^{-1}\frac{\frac{47 g}{20}-\frac{15}{47}}{\frac{460}{47}}\\ &\approx 1.16414700563879\ut{rad}\\ &\approx 1.16\ut{rad}\\ \end{aligned}$$