2-35 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} t_2&=\text{Start}\\ v_4&=0\\ t_8&=\text{Land}\\ g&=9.80665\ut{m/s^2} \end{cases} $$ $$ \begin{aligned} v&=v_0+a\Delta t,\\ 0&=v_0-g(2)\\ \end{aligned} $$ $$\therefore v_0=2g$$ $$\ab{a}$$ $$ \begin{aligned} \Delta y_{\max}&=\int_0^4v\dd t\\ &=\frac{1}{2}\(t_2+t_4\)v_0\\ &=\frac{1}{2}\(2+4\)2g\\ &=6g\\ &=\frac{588399}{10000}\ut{m}\\ &=58.8399\ut{m}\\ &\approx 6\times10\ut{m} \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} \Ans &= -\Delta y_{0\rarr8}\\ &=-\int_0^8v\dd t\\ &=-\(\int_0^4v\dd t+\int_4^8v\dd t\)\\ &=-\(6g+\frac{1}{2}\cdot t\cdot at\)\\ &=-\bra{6g+\frac{1}{2}\cdot 4\cdot (-g)4}\\ &=-\(6g-8g\)\\ &=2g\\ &=\frac{196133}{10000}\ut{m}\\ &=19.6133\ut{m}\\ &\approx2\times10\ut{m} \end{aligned} $$