11판/4. 2차원 운동과 3차원 운동

4-10 할리데이 11판 솔루션 일반물리학

짱세디럭스 2024. 1. 29. 18:24
{h1=10.2[m]v1=(8.90i^+6.70j^)[m/s]g=9.80665[m/s2] \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} (a)\ab{a} 2aS=v2v02,2aS=v^2-{v_0}^2, {2(g)(h1)=(vy1)2vy022(g)(hmax)=(vymax)2vy02 \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} hmax=2gh1(vymax)2+(vy1)22g=2g(10.2)02+(6.7)22g=4489200g+515=12247283980665[m]12.48875304002896[m]12.5[m] \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} (b)\ab{b} 2aS=v2v02,2aS=v^2-{v_0}^2, 2(g)(h1)=(vy1)2vy022(g)(10.2)=(6.7)2vy02 \begin{aligned} 2(-g)(h_1)&=(v_{y1})^2-{v_{y0}}^2\\ 2(-g)(10.2)&=(6.7)^2-{v_{y0}}^2\\ \end{aligned} vy0=2040g+448910v_{y0}=\frac{\sqrt{2040 g+4489}}{10} S=v0t+12at2,Δy=vy0t+12(g)t20=vy0t+12(g)t2=vy012gt \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=2vy0gt=\frac{2v_{y0}}{g} Δx=vxt=(8.9)2vy0g=(8.9)2g2040g+448910=892040g+448950g=71261236415196133[m]28.4075775673929[m]28.4[m] \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} (c,d)\ab{c,d} 2aS=v2v02,2(g)(0)=vy22(2040g+448910)2vy2=2040g+448910v2=8.90i^+2040g+448910j^ \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} (c)\ab{c} v2=(8.9)2+(2040g+448910)2=1.7(12g+73)=1100162077835[m/s]18.00432336967985[m/s]18.0[m/s] \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} (d)\ab{d} θ2=tan12040g+4489108.9=tan1(1892040g+4489)=tan1(122472835890)1.053737352915803[rad]1.05[rad] \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}