4-49 할리데이 11판 솔루션 일반물리학

$$\begin{cases} \vec v_0&=2.5\times10^6\ut{m/s}\\ \Delta x &= 3.3\ut{m}\\ g&=9.80665\ut{m/s^2} \end{cases}$$ $$\ab{a}$$ $$\begin{aligned} t &= \frac{\Delta x}{v_x} \\ &= \frac{3.3}{v_x} \\ \end{aligned}$$ $$\begin{aligned} S&=v_0t+\frac{1}{2}at^2,\\ \Delta y&=(0)t+\frac{1}{2}(-g)t^2\\ &=-\frac{1}{2}g\(\frac{3.3}{v_x}\)^2\\ &=-\frac{1089g}{200{v_x}^2}\\ \end{aligned}$$ $$\begin{aligned} \abs{\Delta y}&=\frac{1089g}{200{v_x}^2}\\ &=8.54355348\times10^{-12}\ut{m}\\ &\approx 8.5\times10^{-12}\ut{m}\\ \end{aligned}$$ $$\ab{b}$$ $$\begin{aligned} \abs{\Delta y}&=\frac{1089g}{200{v_x}^2},\\ \end{aligned}$$ $$\therefore \text{Decrease}$$