솔루션피아

$$\begin{cases} A=\text{Head}\\ B=\text{Body}\\ \end{cases}$$ (a)$v_{A160}=?$ $$\begin{cases} v_0=0\\ a_A = \text{Linear} & \\ a_A(t) = 0 &(t

$$\begin{cases} x_1=0\\ t_1=3.0\ut{s}\\ v_0=-10\ut{m/s}\\ a=-g = -9.80665\ut{m/s^2}\\ \end{cases}$$ $$x_0=?$$ $$\begin{aligned} \Delta x &= v_0t+\frac{1}{2}at^2,\\ x_1-x_0&= (-10)(3.0)+\frac{1}{2}(-g)(3.0)^2\\ x_0&= 30+\frac{9}{2} g\\ &= 30+\frac{9}{2}(9.80665)\\ &=\frac{2965197}{40000}\ut{m}\\ &=74.1299\ut{m}\\ &\approx 74\ut{m}\\ \end{aligned}$$

$$\begin{cases} t_1 = \text{FlowerPot is Bottom of Window}\\ t_2 = \text{FlowerPot is Top of Window}\\ t_3 = \text{FlowerPot is Top Point}\\ \end{cases}$$ $$\begin{cases} x_0=0\\ x_{1\to2}=2.00\ut{m}\\ t_{1\to2}=0.50\ut{s}\\ v_3=0\\ a=-g = -9.80665\ut{m/s^2}\\ \end{cases}$$ $$x_{2\to3}=?$$ $$\Delta x = vt-\frac{1}{2}at^2,$$ $$\Delta x_{1\to2} = v_2t_{1\to2}-\frac{1}{2}(-g)t_{1\to2}^2$$ $$ ..

$$\begin{cases} t_1 = \text{Man is 15cm From Bottom 1st Time}\\ t_2 = \text{Man is Under 15cm From Top 1st Time}\\ t_3 = \text{Man is Top Point}\\ t_4 = \text{Man is Under 15cm From Top 2st Time}\\ t_5 = \text{Man is 15cm From Bottom 2nd Time}\\ t_6 = \text{Man is Landing}\\ \end{cases}$$ $$\begin{cases} x_0=x_6=0\\ x_3=0.780\ut{m}\\ v_3=0\\ x_1=x_5=0.150\ut{m}\\ x_2=x_4=x_3-0.150\ut{m}\\ a=-g..

$$\begin{cases} t_0 = \text{Ball is Top of Building}\\ t_1 = \text{DownBall is Top of Window}\\ t_2 = \text{DownBall is Bottom of Window}\\ t_3 = \text{Ball is Contact Road}\\ t_4 = \text{UpBall is Bottom of Window}\\ t_5 = \text{UpBall is Top of Window}\\ \end{cases}$$ $$\begin{cases} \Delta t_{1\to2}=0.125\ut{s}\\ \Delta t_{4\to5}=0.125\ut{s}\\ \Delta t_{2\to4}=2.00\ut{s}\\ x_0=h\\ x_{1\to2}..

$$\begin{cases} t_1 = \text{Rock is Top of Building}\\ t_2 = \text{Rock is High Point}\\ \end{cases}$$ $$\begin{cases} t_1=1.5\ut{s}\\ t_2=2.5\ut{s}\\ x_0 = 0 \\ x_1 = h \\ v_0 > 0 \\ a=-g = -9.80665\ut{m/s^2}\\ \end{cases}$$ $$h=?$$ $$\Delta x = vt-\frac{1}{2}at^2,$$ $$ \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..