4-24 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} t_1 &= 1.50\ut{s}\\ d&=26.7\ut{m}\\ \theta_1&=60.0\degree\\ \end{cases} $$ $$ \begin{aligned} v_x&=\frac{\Delta x}{t}\\ &=\frac{-26.7}{1.50}\ut{m/s}\\ &=-\frac{89}{5}\ut{m/s}\\ \end{aligned} $$ $$ \begin{aligned} \vec v_1&=v_x\i+v_x\tan\theta_1\j\\ &=\(-\frac{89}{5}\i-\frac{89\sqrt3}{5}\j\)\ut{m/s}\\ \end{aligned} $$ $$\ab{a}$$ $$S=vt-\frac{1}{2}at^2,$$ $$ \begin{aligned} \Delta y&=v_{y1}t-\frac{1}{2}(-g)t^2\\ &=v_{y1}t+\frac{1}{2}gt^2\\ &=\(-\frac{89\sqrt3}{5}\)(1.5)+\frac{1}{2}g(1.5)^2\\ &\approx -35.21327531208902\ut{m}\\ &\approx -35.2\ut{m} \end{aligned} $$ $$h=\Delta y \approx 35.2\ut{m}$$ $$\ab{b,c}$$ $$ \begin{aligned} \vec v&=\vec v_0+\vec at,\\ \vec v_0&=\vec v - at\\ &=\(-\frac{89}{5}\i-\frac{89\sqrt3}{5}\j\)-(-g\j)(1.5)\\ &=-\frac{89}{5}\i+\(\frac{588399}{40000}-\frac{89 \sqrt{3}}{5}\)\j\\ &\approx \(-17.8\i-16.12052937472602\j\)\ut{m/s}\\ &\approx \(-17.8\i-16.1\j\)\ut{m/s}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} v_0&=\sqrt{\(-\frac{89}{5}\)^2+\(\frac{588399}{40000}-\frac{89 \sqrt{3}}{5}\)^2}\\ &\approx 24.01481766163142\ut{m/s}\\ &\approx 24.0\ut{m/s}\\ \end{aligned} $$ $$\ab{c}$$ $$ \begin{aligned} \theta_{0\larr~x+}&=\tan^{-1}\frac{\frac{89 \sqrt{3}}{5}-\frac{588399}{40000}}{\frac{89}{5}}\\ &\approx 0.73592663972889\ut{rad}\\ &\approx 0.736\ut{rad}\\ \end{aligned} $$ $$\ab{d}$$ $$\title{Down}$$