4-17 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} v_0&=22\ut{m/s},\theta_0=60\degree\\ t_1&=1.2\ut{s}\\ \end{cases} $$ $$\ab{a,b,c}$$ $$v=v_0+at,$$ $$ \begin{aligned} v_{y1}&=22\sin60\degree+(-g)(1.2)\\ &=11\sqrt{3}-\frac{6}{5}g\\ \end{aligned} $$ $$ \begin{aligned} \vec v_1&=20\cos60\degree\i+\(11\sqrt{3}-\frac{6}{5}g\)\j\\ &=10\i+\(11\sqrt{3}-\frac{6}{5}g\)\j\\ \end{aligned} $$ $$\ab{a}$$ $$ \begin{aligned} v_1 &= \sqrt{10^2+\(11\sqrt{3}-\frac{6}{5}g\)^2}\\ &\approx 12.37194768443527\ut{m/s}\\ &\approx 12\ut{m/s}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} \theta_1&=\tan^{-1}\frac{11\sqrt{3}-\frac{6}{5}g}{10}\\ &\approx 0.6295710089002262\ut{rad}\\ &\approx 0.63\ut{rad} \end{aligned} $$ $$\ab{c}$$ $$\theta_1>0,$$ $$\title{Up}$$ $$\ab{d,e,f}$$ $$v=v_0+at,$$ $$ \begin{aligned} v_{y3}&=22\sin60\degree+(-g)(3)\\ &=11\sqrt{3}-3g\\ \end{aligned} $$ $$ \begin{aligned} \vec v_3&=20\cos60\degree\i+\(11\sqrt{3}-3g\)\j\\ &=10\i+\(11\sqrt{3}-3g\)\j\\ \end{aligned} $$ $$\ab{d}$$ $$ \begin{aligned} v_1 &= \sqrt{10^2+\(11\sqrt{3}-3g\)^2}\\ &\approx 14.40426320807518\ut{m/s}\\ &\approx 14\ut{m/s}\\ \end{aligned} $$ $$\ab{e}$$ $$ \begin{aligned} \theta_1&=\tan^{-1}\frac{11\sqrt{3}-3g}{10}\\ &\approx -0.8034344094524052\ut{rad}\\ &\approx -0.80\ut{rad} \end{aligned} $$ $$\ab{f}$$ $$\theta_3<0,$$ $$\title{Down}$$