4-26 할리데이 10판 솔루션 일반물리학

$$\begin{cases} \vec v_0 &= (18.0\ut{m/s},40.0\degree)\\ \vec a &= -g\j\\ g &\approx 9.80665\ut{m/s^2}\\ \end{cases}$$ $$\begin{aligned} \vec v_0 &= (18.0\ut{m/s},40.0\degree)\\ &=18\cos40\degree\i+ 18\sin40\degree\j\\ \end{aligned}$$ $$\begin{aligned} \Delta y &= v_{y0}t+\frac{1}{2}a_yt^2,\\ &= 18\sin40\degree t+\frac{1}{2}(-g)t^2\\ &= 18\sin40\degree t-\frac{1}{2}gt^2\\ \end{aligned}$$ $$ \vec r = \(18\cos40\degree t\)\i+\(18\sin40\degree t-\frac{1}{2}gt^2\)\j $$
(a) $\Delta x_{t=1.10}=?$ $$\begin{aligned} \Delta x &= 18\cos40\degree t\\ &= 18\cos40\degree \cdot 1.10\\ &=\frac{99}{5} \cos 40\degree\ut{m}\\ &\approx 15.1676799738\ut{m}\\ &\approx 15.2\ut{m} \end{aligned}$$
(b) $\Delta y_{t=1.10}=?$ $$\begin{aligned} \Delta y &= 18\sin40\degree t-\frac{1}{2}gt^2\\ &= 18\sin40\degree \cdot 1.10-\frac{1}{2}g(1.10)^2\\ &=\frac{99}{5} \sin40 \degree-\frac{121}{200}g\\ &\approx6.79417142179\ut{m}\\ &\approx6.79\ut{m}\\ \end{aligned}$$
(c) $\Delta x_{t=1.80}=?$ $$\begin{aligned} \Delta x &= 18\cos40\degree t\\ &= 18\cos40\degree \cdot 1.80\\ &=\frac{162}{5} \cos40\degree\ut{m}\\ &\approx 24.8198399571\ut{m}\\ &\approx 24.8\ut{m}\\ \end{aligned}$$
(d) $\Delta y_{t=1.80}=?$ $$\begin{aligned} \Delta y &= 18\sin40\degree t-\frac{1}{2}gt^2\\ &= 18\sin40\degree \cdot 1.80-\frac{1}{2}g(1.80)^2\\ &=\frac{162}{5} \sin40\degree-\frac{81}{50}g\\ &\approx4.93954555384\ut{m}\\ &\approx4.94\ut{m}\\ \end{aligned}$$
(e) $\Delta x_{t=5.00}=?$ $$\begin{aligned} \Delta x &= 18\cos40\degree t\\ &= 18\cos40\degree \cdot 5.00\\ &=90 \cos40\degree\ut{m}\\ &\approx 68.9439998807\ut{m}\\ &\approx 68.9\ut{m}\\ \end{aligned}$$
(f) $\Delta y_{t=5.00}=?$ $$\begin{aligned} \Delta y &= 18\sin40\degree t-\frac{1}{2}gt^2\\ &= 18\sin40\degree \cdot 5.00-\frac{1}{2}g(5.00)^2\\ &=90 \sin40\degree-\frac{25}{2}g\\ &\approx -64.7322401282\ut{m}\\ &\approx -64.7\ut{m}\\ \end{aligned}$$