솔루션피아

$$ \begin{cases} v_A&=v\j\\ v_B&=\frac{1}{2}v\j\\ y_B&=y_A+4\ut{m}\\ g&=9.80665\ut{m/s^2} \end{cases} $$ $$\ab{a}$$ $$2aS=v^2-v_0^2,$$ $$ \begin{aligned} 2aS&=v^2-v_0^2\\ 2(-g)(4\ut{m})&=\(\frac{1}{2}v\)^2-(v)^2\\ \end{aligned} $$ $$ \begin{aligned} v&=\sqrt{\frac{32g}{3}}\\ &=\frac{1}{25}\sqrt{\frac{196133}{3}}\ut{m/s}\\ &\approx 10.22762272801782\ut{m/s}\\ &\approx 1\times10\ut{m/s}\\ \end{ali..

$$ \begin{cases} v&=303\ut{km/h}\\ t&=120\ut{ms}\\ \end{cases} $$ $$ \begin{aligned} \Delta x &= vt\\ &=303\ut{km/h}\cdot120\ut{ms}\\ &=303\ut{km/h}\cdot\frac{1000\ut{m}}{1\ut{km}}\cdot\frac{1\ut{h}}{3600\ut{s}}\cdot120\ut{ms}\cdot\frac{1\ut{s}}{1000\ut{ms}}\\ &=\frac{101}{10}\ut{m}\\ &=10.1\ut{m}\\ &\approx 1.0\times10\ut{m} \end{aligned} $$

$$ \begin{cases} t_1 &= 1.12\ut{s} &=\text{Time at }h_{\max}\\ t_2 &= 6.00\ut{s} &=\text{Time at Ground}\\ a&=-g&=-9.80665\ut{m/s^2} \end{cases} $$ $$\ab{a}$$ $$v=v_0+at,$$ $$ \begin{aligned} v_0&=v-at\\ &=0-(-g)t_1\\ &=9.80665\ut{m/s^2}\cdot1.12\ut{s}\\ &=\frac{1372931}{125000}\ut{m/s}\\ &=10.983448\ut{m/s}\\ &\approx11.0\ut{m/s} \end{aligned} $$ $$\ab{b}$$ $$S=vt-\frac{1}{2}at^2,$$ $$ \begin{a..

$$ \begin{cases} a&=-g=-9.80665\ut{m/s^s}\\ v_0&=0\\ t_1&=50\ut{ms}=0.05\ut{s}\\ t_2&=100\ut{ms}=2t_1\\ t_3&=150\ut{ms}=3t_1\\ t_4&=200\ut{ms}=4t_1\\ t_5&=250\ut{ms}=5t_1 \end{cases} $$ $$$$ $$ \begin{aligned} S_k&=v_0t+\frac{1}{2}at^2,\\ &=\frac{1}{2}g(t_k)^2\\ \end{aligned} $$ $$\ab{a}$$ $$ \begin{aligned} S_{1}&=\frac{1}{2}gt_1^2\\ &=\frac{1}{2}(9.80665\ut{m/s^2})(0.05\ut{s})^2\\ &=\frac{1961..

$$ \begin{cases} a&=-g=-9.80665\ut{m/s^2}\\ h&=15\ut{m}\\ v_0&=-10\ut{m/s}\\ \end{cases} $$ $$\ab{a}$$ $$2aS=v^2-v_0^2,$$ $$ \begin{aligned} v&=\sqrt{2as+v_0^2}\\ &=\sqrt{2(-g)(-15)+(-10)^2}\\ &=\sqrt{100+30g}\\ &=\frac{1}{20}\sqrt{\frac{788399}{5}}\ut{m/s}\\ &\approx19.85445793770256\ut{m/s}\\ &\approx19.9\ut{m/s}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} S&=v_0t+\frac{1}{2}at^2\\ -15&=-..

$$ \begin{cases} H&=S_{1.70}\\ h_{M}&=S_{1.70+1.40}=S_{3.10}\\ a&=-g=-9.80665\ut{m/s^2} \end{cases} $$ $$v=v_0+at,$$ $$ \begin{aligned} v_0&=v_{3.1}-at\\ &=(0)-(-g)(3.1)\\ &=3.1g \end{aligned} $$ $$S=v_0t+\frac{1}{2}at^2,$$ $$ \begin{aligned} H&=(3.1g)(1.7)+\frac{1}{2}(-g)(1.7)^2\\ &=\frac{153}{40}g\\ &=\frac{153}{40}\cdot9.80665\ut{m}\\ &=\frac{30008349}{800000}\ut{m}\\ &=37.51043625\ut{m}\\ &\..

$$ \begin{cases} v_{A0}&=60.0\ut{m/s}\\ v_{B0}&=\frac{3}{4}\cdot60.0\ut{m/s}=45.0\ut{m/s}\\ \Delta t_A&=5\ut{s}\\ \Delta t_B&=4\ut{s}\\ v_{A5}&=0\\ v_{B4}&=0\\ \Delta x_{A0\rarr B0}&=320\ut{m} \end{cases} $$ $$S=\frac{1}{2}(v+v_0)t,$$ $$ \begin{aligned} S_A&=\frac{1}{2}(v_{A5}+v_{A0})\Delta t_A\\ &=\frac{1}{2}(0+60.0\ut{m/s})(5\ut{s})\\ &=150\ut{m} \end{aligned} $$ $$ \begin{aligned} S_B&=\frac{..

$$ \begin{cases} v_0&=0\\ v_1&=31\ut{m/s}\\ a_{0\rarr1}&=2.6\ut{m/s^2}\\ a_{1\rarr2}&=-1.4\ut{m/s^2}\\ v_2&=0 \end{cases} $$ $$\ab{a}$$ $$v=v_0+at,$$ $$ \begin{aligned} t_{0\rarr1}&=\frac{v_1-v_0}{a}\\ &=\frac{(31\ut{m/s})-0}{2.6\ut{m/s^2}}\\ &=\frac{155}{13}\ut{s} \end{aligned} $$ $$ \begin{aligned} t_{1\rarr2}&=\frac{v_2-v_1}{a}\\ &=\frac{(0)-(31\ut{m/s})}{-1.4\ut{m/s^2}}\\ &=\frac{155}{7}\ut{..

문제의 의도는 km/h로 추정되며, km/s로 풀 경우 d,e,f의 문항이 성립하지 않습니다. 실제상황으로 생각해도 240km/s는 마하800, 태양계 탈출속도의 14배 정도의 초고속으로, 단순히 풀 경우 문제를 왜곡시킬것이 자명합니다. 따라서 풀이자가 임의로 km/h로 바꾸어 풀겠습니다. $$ \begin{cases} v_0&=240\ut{km/h}=\frac{200}{3}\ut{m/s}\\ S&=174\ut{m}\\ a&=\Cons \end{cases} $$ $$\ab{a}$$ $$2aS=v^2-v_0^2,$$ $$ \begin{aligned} \abs{a}&=\abs{\frac{v^2-v_0^2}{2S}}\\ &=\abs{\frac{(0)^2-(\frac{200}{3}\ut{m/s})^2}{2(17..

$$\ab{a}$$ $$ \begin{aligned} x_5&=x_0+\Delta x_{0\rarr5} \\ &=0+\int_0^5 v\dd t \\ &=\frac{1}{2}(2\ut{s}+4\ut{s})8\ut{m/s}+\frac{1}{2}(4\ut{m/s}+8\ut{m/s})1\ut{s}\\ &=30\ut{m} \end{aligned} $$ $$\ab{b}$$ $$ v_5=4\ut{m/s} $$ $$\ab{c}$$ $$ \begin{aligned} a_5&=\dt v\\ &=\frac{-8\ut{m/s}}{2\ut{s}}\\ &=-4\ut{m/s^2} \end{aligned} $$ $$\ab{d}$$ $$ \begin{aligned} \bar{v}_{1\rarr5}&=\frac{\Delta x_{1\..