4-33 할리데이 10판 솔루션 일반물리학

$$\begin{cases} A : \text{Plane}\\ B : \text{Object} \end{cases}$$ $$\begin{cases} \theta_0 &= 52.0\degree\\ v_A &= \text{Const.}\\ h_0 &= 720\ut{m}\\ \Delta t &= 6.00\ut{s}\\ \end{cases}$$ $$\begin{cases} \vec a &=-g\j\\ g &\approx 9.80665\ut{m/s^2}\\ \end{cases}$$
(a) $v_A=?$ $$\Delta r = v_{0}t+\frac{1}{2}at^2, $$ $$\begin{aligned} \Delta y_B &= v_{By0}t+\frac{1}{2}a_{By}t^2\\ -720 &= v_{By0}\cdot6+\frac{1}{2}(-g)6^2\\ \end{aligned}$$ $$\therefore v_{By0}=3 (g-40)$$ $$\begin{aligned} v_A&=\abs{v_{B0}}\\ &=\abs{\frac{v_{By0}}{\cos\theta}}\\ &=\abs{\frac{3 (g-40)}{\cos52.0\degree}}\\ &\approx147.126389469\ut{m/s}\\ &\approx147\ut{m/s}\\ \end{aligned}$$
(b) $\Delta x_B=?$ $$\begin{aligned} \Delta x_B&=\Delta x_A\\ &=v_{Ax}t\\ &=v_At\sin\theta \\ &=\abs{\frac{3 (g-40)}{\cos52.0\degree}}\cdot 6 \sin52\degree\\ &=18(g-40)\tan52\degree\\ &\approx695.623062247\ut{m}\\ &\approx696\ut{m}\\ \end{aligned}$$
(c) $v_{Bx1}=?$ $$\begin{aligned} v_{Bx1}&=v_{Bx}=v_{Ax}\\ &=\frac{\abs{v_A}}{\sin\theta}\\ &=\frac{\abs{\frac{3 (g-40)}{\cos52.0\degree}}}{\sin52\degree}\\ &\approx 186.706068154\ut{m/s}\\ &\approx 187\ut{m/s}\\ \end{aligned}$$
(d) $v_{By1}=?$ $$\begin{aligned} \Delta r &= vt-\frac{1}{2}at^2, \\ -h_0 &= v_{By1}t-\frac{1}{2}(-g)t^2\\ \end{aligned}$$ $$\begin{aligned} v_{By1}&=-\frac{g t}{2}-\frac{h}{t}\\ &=-\frac{g 6}{2}-\frac{720}{6}\\ &=-120 - 3 g\\ &\approx -149.41995\ut{m/s}\\ &\approx -149\ut{m/s} \end{aligned}$$