10판/2. 직선운동

2-63 할리데이 10판 솔루션 일반물리학

짱세디럭스 2019. 8. 3. 16:27

{t1=FlowerPot is Bottom of Windowt2=FlowerPot is Top of Windowt3=FlowerPot is Top Point \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} {x0=0x12=2.00[m]t12=0.50[s]v3=0a=g=9.80665[m/s2]\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} x23=? x_{2\to3}=? Δx=vt12at2, \Delta x = vt-\frac{1}{2}at^2, Δx12=v2t1212(g)t122 \Delta x_{1\to2} = v_2t_{1\to2}-\frac{1}{2}(-g)t_{1\to2}^2 v2=Δx12t12t12g2=2.000.500.50g2=4g4 \begin{aligned} v_2&=\frac{\Delta x_{1\to2}}{t_{1\to2}}-\frac{t_{1\to2}g}{2}\\ &=\frac{2.00}{0.50}-\frac{0.50g}{2}\\ &=4-\frac{g}{4}\\ \end{aligned} 2aΔx=v2v02,2(g)Δx23=v32v222gΔx23=02v22Δx23=v222g=(4g4)22g=g32+8g1=9.8066532+89.806651=15343033689125525120000[m]=0.122231[m]12.2[cm] \begin{aligned} 2a\Delta x &= v^2-v_0^2,\\ 2(-g)\Delta x_{2\to3} &= v_3^2-v_2^2\\ -2g\Delta x_{2\to3} &= 0^2-v_2^2\\ \Delta x_{2\to3} &=\frac{v_2^2}{2g}\\ &=\frac{(4-\frac{g}{4})^2}{2g}\\ &=\frac{g}{32}+\frac{8}{g}-1\\ &=\frac{9.80665}{32}+\frac{8}{9.80665}-1\\ &=\frac{15343033689}{125525120000}\ut{m}\\ &= 0.122231\ut{m}\\ &\approx 12.2\ut{cm}\\ \end{aligned}