9-29 할리데이 11판 솔루션 일반물리학

$$ \put\begin{cases} 0:\text{Start}\\ 1:\text{Highest Point Before Bomb}\\ 2:\text{Highest Point After Bomb}\\ 3:\text{Landing}\\ \end{cases} $$ $$ \begin{cases} v_0&=24\ut{m/s}\\ \theta_0&=60\degree\\ v_{A1}&=0 \end{cases} $$ $$2aS=v^2-{v_0}^2,$$ $$2(-g)(h)=(0)^2-{v_{y0}}^2$$ $$ \begin{aligned} h_1&=\frac{(v_0\sin\theta)^2}{2g}\taag1\\ \end{aligned} $$ $$ \begin{aligned} v_1&=v_x\\ &=v_0\cos\theta\\ \end{aligned} $$ $$\Delta \Sigma \vec P=0,$$ $$ \begin{aligned} Mv_1&=m_Bv_{B2}\\ \end{aligned} $$ $$ \begin{aligned} v_{B2}&=\frac{2m}{m}v_1\\ &=2v_0\cos\theta\taag2\\ \end{aligned} $$ $$S=v_0t+\frac{1}{2}at^2,$$ $$ \begin{aligned} (-h)&=(0)t+\frac{1}{2}(-g)t^2\\ \end{aligned} $$ $$ \begin{aligned} t&=\sqrt\frac{2h}{g}\\ &=\sqrt{\frac{2}{g}\cdot\frac{(v_0\sin\theta)^2}{2g}}\\ &=\frac{v_0\sin\theta}{g}\taag3\\ \end{aligned} $$ $$ \begin{aligned} \Delta x &= v_x t\\ &=2v_0\cos\theta\cdot\frac{v_0\sin\theta}{g}\\ &=\frac{{v_0}^2}{g}\sin(2\theta)\\ &=\frac{288\sqrt3}{g}\\ &\approx 50.86656835716954\ut{m}\\ &\approx 51\ut{m}\\ \end{aligned} $$