4-53 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} \vec v_{E}&= 8.0\j\ut{m/s}\\ y_0 &= 22\ut{m}\\ \vec v_{0B\larr E}&= 18\j\ut{m/s}\\ g&=9.80665\ut{m/s^2} \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} \vec v_{B0}&=\vec v_{0B\larr E}+\vec v_{E}\\ &=18\j+8.0\j\\ &=26\j\\ \end{aligned} $$ $$ \begin{aligned} 2aS&=v^2-{v_0}^2,\\ 2(-g)(h)&=(0)^2-(v_{By0})^2\\ 2gh&=(18)^2\\ \end{aligned} $$ $$ \begin{aligned} h&=\frac{162}{g}\\ &\approx 16.51940265024244\ut{m}\\ &\approx 17\ut{m} \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} S&=v_0t+\frac{1}{2}at^2,\\ 0&=(18)t+\frac{1}{2}(-g)t^2\\ &=18-\frac{1}{2}gt\\ \end{aligned} $$ $$ \begin{aligned} t&=\frac{36}{g}\\ &\approx 3.670978366720542\ut{s}\\ &\approx 3.7\ut{s}\\ \end{aligned} $$