4-5 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} \Delta y &= 300\ut{m}\\ \vec v_W&=1.8\i\ut{m/s}\\ v_{A\larr W}&=9.0\ut{m/s}\\ \theta_{A\larr W}&=90\degree+30\degree=120\degree\\ \end{cases} $$ $$ \begin{aligned} \vec v_{A\larr W}&=9\cos120\degree\i+9\sin120\degree\j\\ &=-\frac{9}{2}\i+\frac{9}{2}\sqrt3\j\ut{m/s}\\ \end{aligned} $$ $$ \begin{aligned} \vec v_A&=\vec v_{A\larr W}+\vec v_W\\ &=\(-\frac{9}{2}\i+\frac{9}{2}\sqrt3\j\)+1.8\i\\ &=-\frac{27}{10}\i+\frac{9}{2}\sqrt3\j\ut{m/s}\\ \end{aligned} $$ $$\ab{a}$$ $$ \begin{aligned} v_A&=\sqrt{\(-\frac{27}{10}\)^2+\(\frac{9}{2}\sqrt3\)^2}\\ &=\frac{9}{5} \sqrt{21}\ut{m/s}\\ &\approx 8.248636250920512\ut{m/s}\\ &\approx 8.2\ut{m/s} \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} \theta_A&=\tan^{-1}\frac{\frac{9}{2}\sqrt3}{-\frac{27}{10}}\\ &=\pi-\tan ^{-1}\frac{5}{\sqrt3}\\ &\approx 1.9042694990467286\ut{rad}\\ &\approx 1.9\ut{rad}\\ \end{aligned} $$ $$\ab{c}$$ $$ \begin{aligned} t_A&=\frac{\Delta y}{v_{Ay}}\\ &=\frac{300\ut{m}}{\frac{9}{2}\sqrt3\ut{m/s}}\\ &=\frac{200}{3\sqrt3}\ut{s}\\ &\approx 38.49001794597506\ut{s}\\ &\approx 38\ut{s}\\ \end{aligned} $$