4-25 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} A:\text{Wagon}\\ B:\text{Ball}\\ \end{cases} $$ $$ \begin{cases} \vec v_A&=-v_s\i\\ \vec v_{B0\larr A}&=v_{0x}\i+v_{0y}\j\\ g&=9.80665\ut{m/s}\\ \end{cases} $$ $$ \begin{cases} (v_s,\Delta x_{bg})\\ (0,40\ut{m})\\ (10\ut{m/s},0)\\ (20\ut{m/s},-40\ut{m})\\ \end{cases} $$ $$v_{B\larr A}=v_B-v_A,$$ $$ \begin{aligned} \vec v_B&=v_{0x}\i+v_{0y}\j-v_s\i\\ &=\(v_{0x}-v_s\)\i+v_{0y}\j\\ &=v_{Bx}\i+v_{0y}\j\\ \end{aligned} $$ $$ \begin{aligned} \Delta y&=v_{0y}t+\frac{1}{2}(-g)t^2\\ 0&=v_{0y}t-\frac{1}{2}gt^2\\ &=v_{0y}-\frac{1}{2}gt\\ t&=\frac{2v_{0y}}{g} \end{aligned} $$ $$ \begin{aligned} \Delta x &= v_{Bx}t\\ &=\(v_{0x}-v_s\)t \end{aligned} $$ $$ \begin{aligned} \Delta x &=\(v_{0x}-v_s\)\frac{2v_{0y}}{g}\\ g\Delta x &=2v_{0y}\(v_{0x}-v_s\) \end{aligned} $$ $$ \begin{cases} g\Delta x_1 &=2v_{0y}\(v_{0x}-v_{s1}\)\\ g\Delta x_2 &=2v_{0y}\(v_{0x}-v_{s2}\) \end{cases} $$ $$ \begin{cases} g(40) &=2v_{0y}\(v_{0x}-0\)\\ g(0) &=2v_{0y}\(v_{0x}-10\) \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} v_{0x} &= 10\ut{m/s}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} v_{0y} &= 2g\\ &=19.6133\ut{m/s}\\ &\approx 2.0\times10\ut{m/s} \end{aligned} $$ $$\ab{c,d}$$ $$ \begin{aligned} \Delta x_{bs}&=\Delta x_B-\Delta x_A\\ &=v_{Bx}t-v_{Ax}t\\ &=(v_{0x}-v_s)t-(-v_s)t\\ &=v_{0x}t\\ &=v_{0x}\cdot\frac{2v_{0y}}{g}\\ &=(10)\frac{2(2g)}{g}\\ &=40\ut{m} \end{aligned} $$ $$\ab{c}$$ $$40\ut{m}$$ $$\ab{d}$$ $$40\ut{m}$$