4-35 할리데이 10판 솔루션 일반물리학

$$\begin{cases} \Delta x &= 45.7\ut{m}\\ v_0 &= 460\ut{m/s}\\ \Delta y &= 0 \end{cases}$$ $$\begin{cases} \vec a &=-g\j\\ g &\approx 9.80665\ut{m/s^2}\\ \end{cases}$$ $$\begin{cases} t_0 : \text{Start}\\ t_1 : \text{Finish}\\ \end{cases}$$ $$\begin{aligned} \Delta x &= v_x t,\\ &= v_0t\cos\theta\\ \end{aligned}$$ $$t=\frac{\Delta x}{v_0\cos\theta}$$ $$\begin{aligned} S &= v_0 t+ \frac{1}{2}at^2,\\ \Delta y &= v_{y0}t+\frac{1}{2}a_yt^2\\ 0 &=v_0\sin\theta t+\frac{1}{2}(-g)t^2\\ &=v_0\sin\theta -\frac{1}{2}gt\\ &=v_0\sin\theta -\frac{1}{2}g\(\frac{\Delta x}{v_0\cos\theta}\)\\ \end{aligned}$$ $$\theta =\frac{1}{2} \sin ^{-1}\left(\frac{g \Delta x}{v_0^2}\right)$$ $$\begin{aligned} \therefore \theta &\approx 0.00105898922510608\ut{rad}\\ &\approx 1.06\times10^{-3}\ut{rad}\\ &\approx 6.07\times10^{-2}\degree \end{aligned}$$