4-59 할리데이 11판 솔루션 일반물리학

$$\begin{cases} \vec r(0)&=0\i+3.0\j\ut{ft}\\ \vec r(0.25)&=8.0\i+7.0\j\ut{ft}\\ \vec r(0.50)&=16\i+9.0\j\ut{ft}\\ \vec r(0.75)&=24\i+9.0\j\ut{ft}\\ \vec r(1.00)&=32\i+7.0\j\ut{ft}\\ \vec r(1.25)&=40\i+3.0\j\ut{ft}\\ \end{cases}$$ $$\begin{aligned} v_{x0\rarr0.25} &= \frac{8.0\ut{ft}}{0.25\ut{s}}=32\ut{ft/s}\\ v_{x0.25\rarr0.50} &= \frac{8.0\ut{ft}}{0.25\ut{s}}=32\ut{ft/s}\\ v_{x0.50\rarr0.75} &= \frac{8.0\ut{ft}}{0.25\ut{s}}=32\ut{ft/s}\\ v_{x0.75\rarr1.00} &= \frac{8.0\ut{ft}}{0.25\ut{s}}=32\ut{ft/s}\\ v_{x1.00\rarr1.25} &= \frac{8.0\ut{ft}}{0.25\ut{s}}=32\ut{ft/s}\\ \end{aligned}$$ $$\begin{aligned} v_{y0\rarr0.25}&=\frac{+4\ut{ft}}{0.25\ut{s}}=+16\ut{ft/s}\\ v_{y0.25\rarr0.50}&=\frac{+2\ut{ft}}{0.25\ut{s}}=+8\ut{ft/s}\\ v_{y0.50\rarr0.75}&=\frac{0}{0.25\ut{s}}=0\\ v_{y0.75\rarr1.00}&=\frac{-2\ut{ft}}{0.25\ut{s}}=-8\ut{ft/s}\\ v_{y1.00\rarr1.25}&=\frac{-2\ut{ft}}{0.25\ut{s}}=-16\ut{ft/s}\\ \end{aligned}$$ $$\begin{aligned} a_{y0.125\rarr0.375}&=\frac{-8\ut{ft/s}}{0.25\ut{s}}=-32\ut{ft/s^2}\\ a_{y0.375\rarr0.625}&=\frac{-8\ut{ft/s}}{0.25\ut{s}}=-32\ut{ft/s^2}\\ a_{y0.625\rarr0.875}&=\frac{-8\ut{ft/s}}{0.25\ut{s}}=-32\ut{ft/s^2}\\ a_{y0.875\rarr1.125}&=\frac{-8\ut{ft/s}}{0.25\ut{s}}=-32\ut{ft/s^2}\\\end{aligned}$$ $$\therefore \vec a =-32\j\ut{ft/s^2}$$ $$\begin{aligned} \vec v(t) &= \vec v_0+\Delta \vec v\\ &=\vec v_0+\int_0^t\vec a \dd t\\ &=\vec v_0+\int_0^t\(-32\j\) \dd t\\ &=\vec v_0-32t\j \\ \end{aligned}$$ $$\begin{aligned} \vec r(t)&=\vec r_0+\Delta \vec r\\ &=\vec r_0+\int_0^t\vec v\dd t\\ &=\(3\j\)+\int_0^t\(\vec v_0-32t\j\)\dd t\\ &=3\j+\vec v_0t-16t^2\j\\ &=\vec v_0t+(3-16t^2)\j\\ \end{aligned}$$ $$\begin{aligned} \vec v_0&=\frac{\vec r(t)-(3-16t^2)\j}{t}\\ &=\frac{\vec r(1)-(3-16\cdot1^2)\j}{1}\\ &=(32\i+7.0\j)-(-13)\j\\ &=32\i+20\j\\ \end{aligned}$$ $$\ab{a}$$ $$\begin{aligned} v_0&=\sqrt{32^2+20^2}\\ &=4\sqrt{89}\ut{ft/s}\\ &\approx 37.73592452822641\ut{ft/s}\\ &\approx 38\ut{ft/s}\\ \end{aligned}$$ $$\ab{b}$$ $$v_{h\max}=v_x=32\ut{ft/s}$$ $$\ab{c}$$ $$\begin{aligned} 2aS&=v^2-{v_0}^2,\\ 2(-a_y)(\Delta h)&=(0)^2-{v_{y0}}^2\\ \Delta h&=\frac{{v_{y0}}^2}{2a_y}\\ &=\frac{{20}^2}{2\cdot32}\ut{ft}\\ &=\frac{25}{4}\ut{ft}\\ \end{aligned}$$ $$\begin{aligned} H &= h_0+\Delta h\\ &= 3+\frac{25}{4}\ut{ft}\\ &= \frac{37}{4}\ut{ft}\\ &= 9.25\ut{ft}\\ &\approx 9.3\ut{ft}\\ \end{aligned}$$