4-30 할리데이 10판 솔루션 일반물리학

$$\begin{cases} \vec v_{A0} &= (23.1\ut{m/s},45\degree)\\ x_{B0} &= 55\ut{m}\\ \vec a &=-g\j\\ g &\approx 9.80665\ut{m/s^2}\\ \end{cases}$$ $$\bar v_B=?$$ $$\begin{aligned} \vec v_{A0} &= (23.1\ut{m/s},45\degree)\\ &= \frac{231}{10 \sqrt{2}}\i+\frac{231}{10 \sqrt{2}}\j\\ \end{aligned}$$ $$\begin{aligned} \Delta y_A&=v_{Ay0}t+\frac{1}{2}a_yt^2,\\ 0 &= v_{Ay0}t+\frac{1}{2}(-g)t^2\\ &= v_{Ay0}-\frac{1}{2}gt\\ t&=\frac{2v_{Ay0}}{g}\\ &=\frac{231}{5 \sqrt{2} g} \end{aligned}$$ $$\begin{aligned} \Delta x_A&=v_{Ax}t,\\ &=\(\frac{231}{10 \sqrt{2}}\)\(\frac{231}{5 \sqrt{2} g}\)\\ &=\frac{53361}{100 g} \end{aligned}$$ $$\begin{aligned} \bar v_B&=\frac{\Sigma \Delta x_B}{t}\\ &=\frac{55-\Delta x_A}{t}\\ &=\frac{55-\(\frac{53361}{100 g}\)}{\frac{231}{5 \sqrt{2} g}}\\ &=\frac{500 g-4851}{210 \sqrt{2}}\\ &\approx0.176187439646\ut{m/s}\\ &\approx0.18\ut{m/s}\\ \end{aligned}$$