4-44 할리데이 11판 솔루션 일반물리학

$$\begin{cases} v_x&=12.0\ut{m/s}\\g&=9.80665\ut{m/s^2}\\ \end{cases}$$ $$\ab{a}$$ $$\begin{aligned} v&=v_0+at,\\ v_y&=v_{y0}-gt\\ v_y&=0-gt\\ \end{aligned}$$ $$\begin{aligned} v_y(t)&=-gt\\ \vec v(t)&=12\i-gt\j \end{aligned}$$ $$v(t)=\sqrt{12^2+(gt)^2}$$ $$\begin{aligned} v(-1.2)&=\sqrt{12^2+(-1.2g)^2}\\ &\approx 16.80730059469396\ut{m/s}\\ &\approx 16.8\ut{m/s}\\ \end{aligned}$$ $$\ab{b}$$ $$v(t)=\sqrt{12^2+(gt)^2}$$ $$\begin{aligned} v(1.2)&=\sqrt{12^2+(1.2g)^2}\\ &\approx 16.80730059469396\ut{m/s}\\ &\approx 16.8\ut{m/s}\\ \end{aligned}$$ $$\ab{c,d,e,f}$$ $$\vec v=12\i-gt\j,$$ $$\begin{aligned} \vec r(t)&=\vec r_0+\Delta \vec r\\ &=0+\int_0^t\vec v\dd t\\ &=\int_0^t(12\i-gt\j)\dd t\\ &=12t\i-\frac{1}{2}gt^2\j \end{aligned}$$ $$\ab{c}$$ $$\begin{aligned} r_x(-1.2)&=12(-1.2)\\ &=-14.4\ut{m}\\ \end{aligned}$$ $$\ab{d}$$ $$\begin{aligned} r_y(-1.2)&=-\frac{1}{2}g(-1.2)^2\\ &=-0.72g\\ &=-7.060788\ut{m}\\ &\approx -7.06\ut{m}\\ \end{aligned}$$ $$\ab{e}$$ $$\begin{aligned} r_x(1.2)&=12(1.2)\\ &=14.4\ut{m}\\ \end{aligned}$$ $$\ab{f}$$ $$\begin{aligned} r_y(1.2)&=-\frac{1}{2}g(1.2)^2\\ &=-0.72g\\ &=-7.060788\ut{m}\\ &\approx -7.06\ut{m}\\ \end{aligned}$$