2-43 할리데이 11판 솔루션 일반물리학

풀이자 주 : 자유낙하로 가정하지 않으면 문제는 초기속도에 종속된 답이 나오므로, 자유낙하로 가정하고 풀었습니다. $$ \begin{cases} A&:\text{First Half}\\ B&:\text{Second Half}\\ \end{cases} $$ $$ \begin{cases} v_0&=0\\ H&=120\ut{m}\\ h_1&=60\ut{m}\\ g&=9.80665\ut{m/s^2}\\ \end{cases} $$ $$\ab{a}$$ $$ S=v_0t+\frac{1}{2}at^2,$$ $$ \begin{aligned} -h_1&=(0)t+\frac{1}{2}(-g){t_A}^2\\ h_1&=\frac{1}{2}g{t_A}^2\\ 60&=\frac{1}{2}g{t_A}^2\\ \end{aligned} $$ $$ \begin{aligned} t_A&=2 \sqrt{\cfrac{30}{g}}\\ &=400 \sqrt{\frac{15}{196133}}\ut{s}\\ &\approx3.498084412322713\ut{s}\\ &\approx3.5\ut{s}\\ \end{aligned} $$ $$\ab{b}$$ $$ S=v_0t+\frac{1}{2}at^2,$$ $$ \begin{aligned} -H&=(0)t+\frac{1}{2}(-g)t^2\\ H&=\frac{1}{2}gt^2\\ 120&=\frac{1}{2}gt^2\\ t&=4\sqrt{\cfrac{15}{g}}\\ \end{aligned} $$ $$ \begin{aligned} \Ans &= t-t_A\\ &=4\sqrt{\cfrac{15}{g}}-2 \sqrt{\cfrac{30}{g}}\\ &=2\left(2-\sqrt{2}\right)\sqrt{\frac{15}{g}}\\ &=200\left(2-\sqrt{2}\right) \sqrt{\frac{30}{196133}}\ut{s}\\ &\approx 1.448954005909985\ut{s}\\ &\approx 1.4\ut{s}\\ \end{aligned} $$