4-50 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} H:\text{Helicopter}\\ B:\text{Box} \end{cases} $$ $$ \begin{cases} h_{H}&=11.6\ut{m}\\ \vec v_H&=6.20\ut{m/s}\i\\ \vec v_{B0\larr H}&=-15.0\ut{m/s}\i \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} \vec v_{B0}&=\vec v_{B0\larr H}+\vec v_H\\ &=-15.0\ut{m/s}\i+6.20\ut{m/s}\i\\ &=8.8\ut{m/s}\i \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} S&=v_0t+\frac{1}{2}at^2,\\ \Delta y&=(0)t+\frac{1}{2}(-g)t^2\\ -11.6&=-\frac{1}{2}gt^2\\ t&=2\sqrt{\frac{29}{5g}} \end{aligned} $$ $$ \begin{aligned} \Delta x &= v_x t\\ \abs{\Delta x}&=\abs{-15\cdot2\sqrt{\frac{29}{5g}}}\\ &\approx 23.07145125852465\ut{m}\\ &\approx 23.1\ut{m}\\ \end{aligned} $$ $$\ab{c}$$ $$ \begin{aligned} v_x &= v_{B0}= 8.8\ut{m/s} \end{aligned} $$ $$ \begin{aligned} 2aS&=v^2-{v_0}^2,\\ 2(-g)(-11.6)&={v_y}^2-{0}^2\\ v_y&=2\sqrt{\frac{29g}{5}}\\ \end{aligned} $$ $$ \begin{aligned} \theta &=\tan^{-1}\frac{2\sqrt{\frac{29g}{5}}}{8.8}\\ &=\tan^{-1}\frac{\sqrt{145g}}{22}\\ &\approx 1.042660195238836\ut{rad}\\ &\approx 1.04\ut{rad}\\ \end{aligned} $$