10판 187

4-42 할리데이 10판 솔루션 일반물리학

$$\begin{cases} \text{StartPoint}=O&=(0,0)\ut{m}\\ \text{1stTower}=A&=(23,15)\ut{m}\\ \text{2ndTower}=B&=(23+\frac{23}{2},15)\ut{m}\\ \text{Goal}=C&=(R,0)\ut{m}\\ \end{cases}$$ $$\begin{cases} v_0 &= 26.5\ut{m/s}\\ \theta_0 &= 53\degree\\ \end{cases}$$ $$\begin{cases} t_0&=\text{Start Time}\\ t_1&=\text{Over 1stTower Time}\\ t_2&=\text{Over 2ndTower Time}\\ t_3&=\text{Goal Time}\\ \end{cases}$$ ..

4-41 할리데이 10판 솔루션 일반물리학

$$\begin{cases} \phi &= 36.0\degree\\ d &= 0.900\ut{m}\\ v_0 &= 3.56\ut{m/s} \end{cases}$$ $$\begin{cases} \vec a &=-g\j\\ g &\approx 9.80665\ut{m/s^2}\\ \end{cases}$$ $$H = d\sin\phi$$ $$\begin{aligned} 2a\Delta y&=v_y^2-v_{0y}^2,\\ 2(-g)(d\sin\phi)&=(0)^2-(v_0\sin\theta_0)^2\\ \sin\theta_0&=\frac{\sqrt{2dg\sin\phi} }{v_0}\\ \end{aligned}$$ $$\begin{aligned} \theta_0&=\sin^{-1}\frac{\sqrt{2dg\s..

4-40 할리데이 10판 솔루션 일반물리학

$$\begin{cases} H &= 2.160\ut{m}\\ v_0 &= 15.00\ut{m/s}\\ \end{cases}$$ $$\begin{cases} \vec a &=-g\j\\ g &\approx 9.80665\ut{m/s^2}\\ \end{cases}$$ $$\begin{aligned} \Delta y&=v_{0y}t+\frac{1}{2}at^2\\ -H&=v_0t\sin\theta-\frac{1}{2}gt^2\\ -(2.160)&=(15.00)t\sin\theta-\frac{1}{2}gt^2\\ \end{aligned}$$ $$ \therefore t=\frac{3}{5 g} \left(\sqrt{12 g+625 \sin ^2\theta}+25 \sin \theta\right)$$ $$\be..

4-39 할리데이 10판 솔루션 일반물리학

$$\begin{cases} d_{1.50} &= -25.0\ut{m}\\ \theta_{1.50} &= 60.0\degree\\ \end{cases}$$ $$\begin{cases} \vec a &=-g\j\\ g &\approx 9.80665\ut{m/s^2}\\ \end{cases}$$ (a) $h_0=?$ $$\begin{aligned} d&=v_x t\\ v_x&=\frac{25}{1.5}\\ &= \frac{50}{3} \end{aligned}$$ $$\begin{aligned} \vec v_{1.5}&=(-v_{x1.5},-v_{y1.5})\\ &=(-v_x,-\sqrt{3}v_x)\\ &=\(-\frac{50}{3},-\frac{50}{\sqrt{3}}\)\\ \end{aligned}$$ ..

4-38 할리데이 10판 솔루션 일반물리학

$$\begin{cases} t_0 : \text{Start}, 0\ut{sec}\\ \end{cases}$$ $$\begin{cases} v_0 &= 31\ut{m/s}\\ v_{2.5} &= 19\ut{m/s}\\ v_5 &= 31\ut{m/s}\\ \end{cases}$$ $$\begin{cases} \vec a &=-g\j\\ g &\approx 9.80665\ut{m/s^2}\\ \end{cases}$$ (a) $\Delta x_{0\rarr\text{Ground}}=?$ $$\begin{aligned} \abs{\vec v}_{\min}&=\vec v_{\text{HighMax}},\\ t_{2.5}&=t_{\text{HighMax}}\\ \end{aligned}$$ $$\begin{align..

4-37 할리데이 10판 솔루션 일반물리학

$$\begin{cases} y_0 &= 12\ut{m}\\ v_x &= 2.50\ut{m/s}\\ \end{cases}$$ $$\begin{cases} \vec a &=-g\j\\ g &\approx 9.80665\ut{m/s^2}\\ \end{cases}$$ $$\begin{cases} t_0 : \text{Start}\\ t_1=0.900\ut{sec}\\ t_2 : \text{Finish}\\ \end{cases}$$ (a) $\Delta x_1=?$ $$\begin{aligned} \Delta x &= v_xt,\\ &=2.25\ut{m} \end{aligned}$$ (b) $y_1=?$ $$\begin{aligned} S&=v_0t+\frac{1}{2}at^2,\\ \Delta y &= v_{..

4-36 할리데이 10판 솔루션 일반물리학

$$\begin{cases} y_0 &= 2.42\ut{m}\\ v_x &= 23.6\ut{m/s}\\ h_N &= 0.90\ut{m}\\ \Delta x &= 12\ut{m} \end{cases}$$ $$\begin{cases} \vec a &=-g\j\\ g &\approx 9.80665\ut{m/s^2}\\ \end{cases}$$ $$\begin{cases} t_0 : \text{Start}\\ t_1 : \text{On Net}\\ \end{cases}$$ (a),(b) $\text{horizon}$ $$\Delta x = v_x t,$$ $$\therefore t = \frac{\Delta x}{v_x}$$ $$\begin{aligned} S &= v_0t+\frac{1}{2}at^2,\\ \..

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,..

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}\\ &..

4-33 할리데이 10판 솔루션 일반물리학

$$\begin{cases} A : \text{Plane}\\ B : \text{Object} \end{cases}$$ $$\begin{cases} \theta_0 &= 52.0\degree\\ v_A &= \text{Const.}\\ h_0 &= 720\ut{m}\\ \Delta t &= 6.00\ut{s}\\ \end{cases}$$ $$\begin{cases} \vec a &=-g\j\\ g &\approx 9.80665\ut{m/s^2}\\ \end{cases}$$ (a) $v_A=?$ $$\Delta r = v_{0}t+\frac{1}{2}at^2, $$ $$\begin{aligned} \Delta y_B &= v_{By0}t+\frac{1}{2}a_{By}t^2\\ -720 &= v_{By0}..