10판/2. 직선운동

2-61 할리데이 10판 솔루션 일반물리학

짱세디럭스 2019. 8. 2. 22:16

{t0=Ball is Top of Buildingt1=DownBall is Top of Windowt2=DownBall is Bottom of Windowt3=Ball is Contact Roadt4=UpBall is Bottom of Windowt5=UpBall is Top of Window \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} {Δt12=0.125[s]Δt45=0.125[s]Δt24=2.00[s]x0=hx12=x45=1.20[m]v0=0v1=v5v2=v4a=g=9.80665[m/s2]\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}=-x_{4\to5}=-1.20\ut{m}\\ v_0=0\\ v_1=-v_5\\ v_2=-v_4\\ a=-g = -9.80665\ut{m/s^2}\\ \end{cases} h=? h=? t23=t34t_{2\to3}=t_{3\to4} (Same Motion in 1-3 and 3-5) (\because \text{Same Motion in 1-3 and 3-5}) t23=12t24=1.00[s]t_{2\to3}=\frac{1}{2}t_{2\to4}=1.00\ut{s} Δx=v0t+12at2, \Delta x = v_0t+\frac{1}{2}at^2, {Δx01=v0t01+12(g)t012Δx02=v0t02+12(g)t022\begin{cases} \Delta x_{0\to1} &= v_0t_{0\to1}+\frac{1}{2}(-g)t_{0\to1}^2\\ \Delta x_{0\to2} &= v_0t_{0\to2}+\frac{1}{2}(-g)t_{0\to2}^2\\ \end{cases} {x1x0=12g(t1t0)2x2x0=12g(t2t0)2\begin{cases} x_1-x_0 &= -\frac{1}{2}g(t_1-t_0)^2\\ x_2-x_0 &= -\frac{1}{2}g(t_2-t_0)^2\\ \end{cases} {x1h=12gt12x2h=12gt22\begin{cases} x_1-h &= -\frac{1}{2}gt_1^2\\ x_2-h &= -\frac{1}{2}gt_2^2\\ \end{cases} x2x1=12g(t22t12)x12=12g(t1+t12)2t12=12gt12(2t1+t12) \begin{aligned} x_2-x_1 &= -\frac{1}{2}g(t_2^2-t_1^2)\\ x_{1\to2} &= -\frac{1}{2}g\bra{(t_1+t_{1\to2})^2-t_1^2}\\ &= -\frac{1}{2}gt_{1\to2}(2t_1+t_{1\to2})\\ \end{aligned} t1=x12t12gt122\therefore t_1=-\frac{x_{1\to2}}{t_{1\to2}g}-\frac{t_{1\to2}}{2} t3=t1+t12+t23=x12t12gt122+t12+t23=t122x12t12g+t23=0.12521.200.125g+1.00=485g+1716 \begin{aligned} t_3&=t_1+t_{1\to2}+t_{2\to3}\\ &=-\frac{x_{1\to2}}{t_{1\to2}g}-\frac{t_{1\to2}}{2}+t_{1\to2}+t_{2\to3}\\ &=\frac{t_{1\to2}}{2}-\frac{x_{1\to2}}{t_{1\to2}g}+t_{2\to3}\\ &=\frac{0.125}{2}-\frac{-1.20}{0.125g}+1.00\\ &=\frac{48}{5 g}+\frac{17}{16} \end{aligned} Δx=v0t+12at2,Δx03=v0t03+12(g)t032h=12gt032h=12g(485g+1716)2=(85g+768)212800g=(859.80665+768)2128009.80665=410401800001212008401920000[m]20.43424654768354[m] \begin{aligned} \Delta x &= v_0t+\frac{1}{2}at^2,\\ \Delta x_{0\to3} &= v_0t_{0\to3}+\frac{1}{2}(-g)t_{0\to3}^2\\ -h &= -\frac{1}{2}gt_{0\to3}^2\\ h &= \frac{1}{2}g\(\frac{48}{5 g}+\frac{17}{16}\)^2\\ &=\frac{(85 g+768)^2}{12800 g}\\ &=\frac{(85 \cdot9.80665+768)^2}{12800 \cdot 9.80665}\\ &=\frac{41040180000121}{2008401920000}\ut{m}\\ &\approx 20.43424654768354\ut{m} \end{aligned}