4-34 할리데이 10판 솔루션 일반물리학

$$\begin{cases} v_0 &= 30.0\ut{m/s}\\ \theta_0 &= 40.0\degree\\ \end{cases}$$ $$\begin{cases} \vec a &=-g\j\\ g &\approx 9.80665\ut{m/s^2}\\ \end{cases}$$ $$\begin{cases} t_0 : \text{Start}\\ t_1 : \text{Vertical Max}\\ t_2 : \text{Vertical Max after half point}\\ \end{cases}$$
(a) $v_1=?$ $$\begin{aligned} v_1 &= v_x\\ &= v_0\cos\theta_0\\ &= 30\cos40\degree\\ &\approx 22.9813332936\ut{m/s}\\ &\approx 23.0\ut{m/s}\\ \end{aligned}$$
(b) $v_2=?$ $$\begin{aligned} v_{x2}&=v_x\\ &=v_0\cos\theta_0\\ \end{aligned}$$ $$\begin{aligned} 2as &= v^2-v_0^2,\\ 2(-g)h_{\max} &= 0^2-v_{y0}^2\\ \end{aligned}$$ $$\begin{aligned} h_{\max}&=\frac{v_{y0}^2}{2 g} \end{aligned}$$ $$\begin{aligned} 2as &= v^2-v_0^2,\\ 2(-g)\(\frac{h_{\max}}{2}\)&=v_{y2}^2-v_{y0}^2\\ \end{aligned}$$ $$\begin{aligned} v_{y2}^2&=v_{y0}^2-gh_{\max}\\ &=v_{y0}^2-g\frac{v_{y0}^2}{2 g}\\ &=\frac{1}{2}(v_0\sin\theta_0)^2\\ \end{aligned}$$ $$\begin{aligned} v_2 &= \sqrt{v_{x2}^2+v_{y2}^2}\\ &= \sqrt{(v_0\cos\theta_0)^2+\frac{1}{2}(v_0\sin\theta_0)^2}\\ &=\frac{v_0}{2}\sqrt{ \cos (2 \theta_0 )+3}\\ &\approx 26.7221039586\ut{m/s}\\ &\approx 26.7\ut{m/s}\\ \end{aligned}$$
(c) $v_2/v_1=?$ $$\begin{aligned} \frac{v_2}{v_1}&=\frac{\frac{v_0}{2}\sqrt{ \cos (2 \theta_0 )+3}}{v_0\cos\theta_0}\\ &\approx1.1627743098 \end{aligned}$$ $$\therefore 16.3\%$$