솔루션피아

(풀이자주 : z축의 포물선 운동을 고려하면 이 문제는 풀리지 않습니다. 문제에서 지적하는 화차의 너비 정보가 필수적으로 필요합니다. 따라서 z축의 거동은 무시하고 풀겠습니다.) $$ \begin{cases} A:\text{Bullet}\\ B:\text{Train}\\ \end{cases} $$ $$ \begin{cases} v_B&=110\ut{km/h}=\cfrac{275}{9}\ut{m/s}\\ v_{A0}&=600\ut{m/s}\\ v_{A1}&=0.6\cdot600\ut{m/s}=360\ut{m/s}\\ g&=9.80665\ut{m/s^2} \end{cases} $$ $$ \begin{aligned} \vec v_{A1}&=v_{Ax}\i+v_{Ay}\j,\\ \end{aligned} $$ $..

$$ \begin{cases} v_0&=35.6\ut{m/s}\\ R&=26.4\ut{m}\\ g&=9.80665\ut{m/s^2} \end{cases} $$ $$ \begin{aligned} S&=v_0t+\frac{1}{2}at^2,\\ \Delta y &=v_{y0}t+\frac{1}{2}(-g)t^2\\ 0&=v_{y0}t-\frac{1}{2}gt^2\\ &=v_{y0}-\frac{1}{2}gt\\ \end{aligned} $$ $$t=\frac{2v_{y0}}{g}=\frac{2v_0\sin\theta}{g}$$ $$ \begin{aligned} \Delta x &= v_x t\\ R &= (v_0\cos\theta)\(\frac{2v_0\sin\theta}{g}\)\\ &=\frac{{v_0}..

$$ \begin{cases} \theta_1=30.0\degree\\ \theta_2=50.0\degree\\ \theta_3=80.0\degree\\ d_1=5.00\ut{m}\\ d_2=9.20\ut{m}\\ d_3=13.5\ut{m}\\ \end{cases} $$ $$ \begin{aligned} \phi_2 &= 180\degree-(\theta_2-\theta_1)\\ &=180\degree-(50\degree-30\degree)\\ &=160\degree\\ \end{aligned} $$ $$ \begin{aligned} \phi_3 &= \phi_2-\theta_3+180\degree\\ &=160\degree-80\degree+180\degree\\ &=260\degree \end{ali..

$$ \begin{cases} H&=2.24\ut{m}\\ a&=9.0\ut{m}\\ x_0&=-7.0\ut{m}\\ y_0&=3.0\ut{m}\\ \vec v_0&=v_0\i \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} S&=v_0t+\frac{1}{2}at^2,\\ \Delta y&=(0)t+\frac{1}{2}(-g)t^2\\ -3&=-\frac{1}{2}gt^2\\ t&=\sqrt{\frac{6}{g}}\\ \end{aligned} $$ $$ \begin{aligned} \Delta x &= v_x t\\ 7 &= v_x \sqrt{\frac{6}{g}}\\ \end{aligned} $$ $$ \begin{aligned} v_x &=7\sqrt{\frac{g}{..

$$ \begin{cases} R&=5.00\times10^6\ut{m}\\ T&=22.5\ut{h}=81000\ut{s} \end{cases} $$ $$\ab{a}$$ $$v=\frac{2\pi R}{T},$$ $$ \begin{aligned} a_R&=\frac{v^2}{R}\\ &=\frac{1}{R}\(\frac{2\pi R}{T}\)^2\\ &=\frac{4\pi^2 R}{T^2}\\ &=\frac{4\pi^2 (5.00\times10^6\ut{m})}{(81000\ut{s})^2}\\ &=\frac{20 \pi ^2}{6561}\ut{m/s^2}\\ &\approx 0.03008567109004529\ut{m/s^2}\\ &\approx 0.0301\ut{m/s^2}\\ \end{aligned..

$$ \begin{cases} A:\text{Ball}\\ B:\text{Floor}\\ C:\text{Elevator}\\ \end{cases} $$ $$ \begin{cases} v_{0A\larr B}&=v_0=8.00\ut{m/s}\\ v_{C}&=2.00\ut{m/s}\\ g&=9.80665\ut{m/s^2}\\ \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} 2aS&=v^2-{v_0}^2,\\ 2(-g)(H)&=(0)^2-{8}^2\\ \end{aligned} $$ $$ \begin{aligned} H&=\frac{32}{g}\\ &\approx 3.263091881529371\ut{m}\\ &\approx 3.26\ut{m}\\ \end{aligned} $$ ..

$$ \begin{cases} R&=4.00\ut{m}\\ T&=20.0\ut{s} \end{cases} $$ $$ \begin{aligned} \vec O &= R\j\\ &= 4\j \end{aligned} $$ $$ \begin{aligned} \omega &= \frac{2\pi}{T}\\ &=\frac{\pi}{10}\ut{rad/s} \end{aligned} $$ $$ \begin{aligned} \vec R &= R\cos\theta\i+R\sin\theta\j\\ &= R\cos(\theta_0+\Delta\theta)\i+R\sin(\theta_0+\Delta\theta)\j\\ &= R\cos(\theta_0+\omega t)\i+R\sin(\theta_0+\omega t)\j\\ &=..

$$ \begin{cases} \theta&=70\degree\\ d_1 &= 1.0\ut{ft}\\ d_2 &= 14\ut{ft}\\ h_1 &= 7.0\ut{ft}\\ h_2 &= 10\ut{ft}\\ g&=9.80665\ut{m/s^2} \end{cases} $$ $$381\ut{m}=1250\ut{ft},$$ $$ \begin{aligned} g&=\frac{196133}{6096}\ut{ft/s^2}\\ &\approx 32.17404855643045\ut{ft/s^2}\\ &\approx 32\ut{ft/s^2} \end{aligned} $$ $$ \begin{cases} \Delta x &=d_2-d_1= 13\ut{ft}\\ \Delta y &=h_2-h_1=3.0\ut{ft}\\ \end..

$$ \begin{cases} h&=6.20\ut{m}\\ v_0&=13.6\ut{m/s}\\ g&=9.80665\ut{m/s^2}\\ \end{cases} $$ $$ \begin{aligned} 2aS&=v^2-{v_0}^2,\\ 2(-g)(h)&=(0)^2-{v_{y0}}^2\\ v_{y0}&=\sqrt{2gh}\\ \end{aligned} $$ $$\ab{a}$$ $$ \begin{aligned} \theta &= \sin^{-1}\frac{v_{y0}}{v_0}\\ &= \sin^{-1}\frac{\sqrt{2gh}}{13.6}\\ &= \sin^{-1}\frac{\sqrt{2\cdot6.2g}}{13.6}\\ &\approx 0.9455767031562586\ut{rad}\\ &\approx 0..

$$ \begin{cases} t&=4.70\ut{s}\\ \Delta x &= 46\ut{m}\\ y_0&=150\ut{cm}\\ g&=9.80665\ut{m/s^2} \end{cases} $$ $$ \begin{aligned} v_x &= \frac{\Delta x}{t}\\ &= \frac{46}{4.70}\ut{m/s}\\ &= \frac{460}{47}\ut{m/s}\\ \end{aligned} $$ $$ \begin{aligned} S&=v_0t+\frac{1}{2}at^2,\\ \Delta y&=v_0t+\frac{1}{2}(-g)t^2\\ -1.5\ut{m}&=v_{y0}(4.7)+\frac{1}{2}(-g)(4.7)^2\\ v_{y0}&=\frac{47 g}{20}-\frac{15}{47..