4-10 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} h_1&=10.2\ut{m}\\ \vec v_1&=\(8.90\i+6.70\j\)\ut{m/s}\\ g&=9.80665\ut{m/s^2}\\ \end{cases} $$ $$\ab{a}$$ $$2aS=v^2-{v_0}^2,$$ $$ \begin{cases} 2(-g)(h_1)&=(v_{y1})^2-{v_{y0}}^2\\ 2(-g)(h_{\max})&=(v_{y\max})^2-{v_{y0}}^2\\ \end{cases} $$ $$ \begin{aligned} h_{\max}&=\frac{2 g h_1-(v_{y\max})^2+(v_{y1})^2}{2 g}\\ &=\frac{2 g (10.2)-0^2+(6.7)^2}{2 g}\\ &=\frac{4489}{200 g}+\frac{51}{5}\\ &=\frac{12247283}{980665}\ut{m}\\ &\approx 12.48875304002896\ut{m}\\ &\approx 12.5\ut{m}\\ \end{aligned} $$ $$\ab{b}$$ $$2aS=v^2-{v_0}^2,$$ $$ \begin{aligned} 2(-g)(h_1)&=(v_{y1})^2-{v_{y0}}^2\\ 2(-g)(10.2)&=(6.7)^2-{v_{y0}}^2\\ \end{aligned} $$ $$v_{y0}=\frac{\sqrt{2040 g+4489}}{10} $$ $$ \begin{aligned} S&=v_0t+\frac{1}{2}at^2,\\ \Delta y &=v_{y0}t+\frac{1}{2}(-g)t^2\\ 0 &=v_{y0}t+\frac{1}{2}(-g)t^2\\ &=v_{y0}-\frac{1}{2}gt\\ \end{aligned} $$ $$t=\frac{2v_{y0}}{g}$$ $$ \begin{aligned} \Delta x &= v_x t\\ &=(8.9)\frac{2v_{y0}}{g}\\ &=(8.9)\cdot\frac{2}{g}\cdot\frac{\sqrt{2040 g+4489}}{10}\\ &=\frac{89 \sqrt{2040 g+4489}}{50 g}\\ &=\frac{712 \sqrt{61236415}}{196133}\ut{m}\\ &\approx 28.4075775673929\ut{m}\\ &\approx 28.4\ut{m}\\ \end{aligned} $$ $$\ab{c,d}$$ $$ \begin{aligned} 2aS&=v^2-{v_0}^2,\\ 2(-g)(0)&={v_{y2}}^2-\(\frac{\sqrt{2040 g+4489}}{10}\)^2\\ v_{y2}&=\frac{\sqrt{2040 g+4489}}{10}\\ \vec v_2&=8.90\i+\frac{\sqrt{2040 g+4489}}{10}\j\\ \end{aligned} $$ $$\ab{c}$$ $$ \begin{aligned} v_2 &= \sqrt{(8.9)^2+\(\frac{\sqrt{2040 g+4489}}{10}\)^2}\\ &=\sqrt{1.7(12 g+73)}\\ &=\frac{1}{100}\sqrt{\frac{16207783}{5}}\ut{m/s}\\ &\approx 18.00432336967985\ut{m/s}\\ &\approx 18.0\ut{m/s}\\ \end{aligned} $$ $$\ab{d}$$ $$ \begin{aligned} \theta_2&=\tan^{-1}\frac{\frac{\sqrt{2040 g+4489}}{10}}{8.9}\\ &=\tan ^{-1}\left(\frac{1}{89} \sqrt{2040 g+4489}\right)\\ &=\tan ^{-1}\left(\frac{\sqrt{\frac{12247283}{5}}}{890}\right)\\ &\approx 1.053737352915803\ut{rad}\\ &\approx 1.05\ut{rad}\\ \end{aligned} $$