솔루션피아

$$ \begin{cases} t_1 = \text{Rock is Top of Building}\\ t_2 = \text{Rock is High Point}\\ \end{cases} $$ $$\begin{cases} t_1=1.5\ut{s}\\ t_2=2.5\ut{s}\\ x_0 = 0 \\ x_1 = h \\ v_0 > 0 \\ a=-g = -9.80665\ut{m/s^2}\\ \end{cases} $$ $$ h=?$$ $$ \Delta x = vt-\frac{1}{2}at^2,$$ $$ \begin{aligned} \Delta x_{0\to2} &= v_2t_{0\to2}-\frac{1}{2}(-g)t_{0\to2}^2\\ (x_2-x_0) &= v_2(t_2-t_0)+\frac{1}{2}gt(t_2..

$$ \begin{cases} d = \text{Water Drop Interval Time}\\ A,B,C,D = \text{1,2,3,4st Water}\\ t_N = \text{Nst Water Start}\\ t_4 = \text{1st Water Landing}\\ \end{cases} $$ $$\begin{cases} h=x_{A1}=x_{B2}=x_{C3}=x_{D4}=0.2\ut{m}\\ x_{A4}=0\\ t_1 = 0\\ t_2 = d\\ t_3 = d+t_2=2d\\ t_4 = d+t_3=3d\\ v_0=v_{A1}=v_{B2}=v_{C3}=v_{D4}=0\\ a=-g = -9.80665\ut{m/s^2}\\ \end{cases} $$ $$ \begin{aligned} \Delta x..

$$ \begin{cases} t_1 = \text{Landing Before 1 second}\\ t_2 = \text{Landing}\\ \end{cases} $$ $$\begin{cases} t_{1\to2} = 1\ut{s}\\ x_0 = h\\ x_1 = 0.6h\\ x_2 = 0\\ v_0 = 0\\ a=-g = -9.80665\ut{m/s^2}\\ \end{cases} $$ $$ t_2-t_1 = 1\ut{s},$$ $$ \therefore t_1 = t_2-1\ut{s}$$ $$\Delta x = v_0t+\frac{1}{2}at^2,$$ $$\title{Make System of Equations}$$ $$\begin{cases} \Delta x_{0\to1} &= v_0t_{0\to1}..

$$ \begin{cases} t_1 = \text{Contact Start Floor}\\ t_2 = \text{Contact End Floor}\\ t_3 = \text{ReUprising HighPoint}\\ \end{cases} $$ $$ \begin{aligned} t_{1\to2} &= 12.0\ut{ms}\\ &= 12.0\ut{ms}\cdot\frac{1\ut{s}}{1000\ut{ms}}\\ &= \frac{3}{250}\ut{s}\\ \end{aligned} $$ $$\begin{cases} x_0 = 4.00\ut{m}\\ x_3 = 2.00\ut{m}\\ x_1 = x_2 = 0\\ v_0 = v_3 = 0\\ t_{1\to2} = \frac{3}{250}\ut{s}\\ a=-g ..

$$\begin{cases} v_0 > 0 \\ v_B = \frac{v_A}{3}\\ \Delta y_{A \to B}=0.40\ut{m}\\ a=-g = -9.80665\ut{m/s^2}\\ \end{cases} $$ $$v_A =?$$ $$ 2a\Delta x = v^2-v_0^2,$$ $$ \begin{aligned} 2(-g)\Delta y_{A\to B} &= v_B^2-v_A^2\\ -2g\Delta y_{A\to B} &= \(\frac{v_A}{3}\)^2-v_A^2\\ &= -\frac{8}{9}v_A^2\\ \end{aligned} $$ $$ \begin{aligned} v_A^2 &= \frac{9}{4}g\Delta y_{A\to B}\\ v_A &= \sqrt{\frac{9}{4..

$$\begin{cases} t_0 = \text{Rock Fall Start}\\ t_1 = \text{Rock Contact Start}\\ t_2 = \text{Rock Stop}\\ \end{cases}$$ $$ \begin{aligned} t_{1\to2} &= 20.0\ut{ms}\\ &=20.0\ut{ms}\cdot\frac{1\ut{s}}{1000\ut{ms}}\\ &=\frac{1}{50}\ut{s} \\ \end{aligned} $$ $$\begin{cases} x_0 = 15.0\ut{m}\\ x_1 = x_2=0\ut{m}\\ t_{1\to2} = \frac{1}{50}\ut{s}\\ v_0 = 0\\ v_2 = 0\\ a=-g = -9.80665\ut{m/s^2}\\ \end{ca..

$$\begin{cases} A = \text{1st Rock}\\ B = \text{2st Rock}\\ t_0 = \text{1st Rock Start}\\ t_1 = \text{2nd Rock Start}\\ t_2 = \text{A,B Rocks Landing}\\ \end{cases}$$ $$\begin{cases} x_{A0} = x_{B0}=x_{B1}=x_0=53.6\ut{m}\\ x_{A2} = x_{B2}=x_2=0\ut{m}\\ t_1 = 1.00\ut{s}\\ v_{A0} = 0\\ a=a_A = a_B =-g = -9.80665\ut{m/s^2}\\ \end{cases} $$ (a)$v_{B1}=?$ $$ \Delta x = v_0t+\frac{1}{2}at^2,$$ $$ \beg..

$$\begin{cases} K = \text{Key}\\ B = \text{Boat}\\ \end{cases}$$ $$\begin{cases} x_{K0} = 45\ut{m}\\ x_{K1} = 0\ut{m}\\ v_{K0} = 0\\ a_K = -g = -9.80665\ut{m/s^2}\\ \end{cases} $$ $$ \begin{cases} x_{B0} = -12\ut{m}\\ x_{B1} = 0\ut{m}\\ v_B = \Cons\\ t=t_{K0\to1} = t_{B0\to1}\\ \end{cases}$$ $$v_B=?$$ $$ \Delta x = v_0t+\frac{1}{2}at^2,$$ $$ \Delta x_{K0\to1} = \frac{1}{2}(-g)t^2$$ $$ t^2=\frac{..

$$\begin{cases} t_{1} = \text{Last 20\% Point}\\ t_{2} = \text{Landing}\\ \end{cases}$$ $$\begin{cases} x_0 = 90\ut{m}\\ x_1 = \frac{20}{100}x_0=\frac{1}{5}x_0\\ x_2 = 0\\ v_0 = 0\\ a = -g = -9.80665\ut{m/s^2}\\ \end{cases}$$ (a)$t_{1\to2}=?$ $$\Delta x = v_0t+\frac{1}{2}at^2,$$ $$\Delta x_{0\to t} = \frac{1}{2}(-g)t^2$$ $$\therefore t=\sqrt{-\frac{2(x_t-x_0)}{g} }$$ $$ \begin{aligned} t_1&=\sqr..

(풀이자주 : 정황상 가속도가 자연스럽게 변하여 0에 도달했다고 읽히지만, 본 챕터에서 다루는 범위를 넘어서는 관계로 분절된 가속도를 가진 문제로 간주하고 풀었습니다. 자연스럽게 가속도가 변해 종단속도에 도달했다면 답은 달라집니다.) $$\begin{cases} t_{1} = \text{Middle Point}\\ t_{2} = \text{Landing}\\ \end{cases}$$ $$\begin{cases} x_0 = 39.2\ut{m}\\ x_1 = \frac{1}{2}x_0\\ x_2 = 0\\ v_0 = 0\\ a_{0\to1} = -g = -9.80665\ut{m/s^2}\\ a_{1\to2} = 0\\ \end{cases}$$ $$v_2=?$$ $$a_{1\to2} = 0,$$ $$\ther..