10판/2. 직선운동

2-62 할리데이 10판 솔루션 일반물리학

짱세디럭스 2019. 8. 2. 23:56

{t1=Man is 15cm From Bottom 1st Timet2=Man is Under 15cm From Top 1st Timet3=Man is Top Pointt4=Man is Under 15cm From Top 2st Timet5=Man is 15cm From Bottom 2nd Timet6=Man is Landing \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} {x0=x6=0x3=0.780[m]v3=0x1=x5=0.150[m]x2=x4=x30.150[m]a=g=9.80665[m/s2]\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 = -9.80665\ut{m/s^2}\\ \end{cases} 2aΔx=v2v02, 2a\Delta x = v^2-v_0^2, 2(g)Δx03=v32v022gx3=v02v02=2gx3v0=2gx3 (v0>0) \begin{aligned} 2(-g)\Delta x_{0\to3} &= v_3^2-v_0^2\\ -2gx_3 &= -v_0^2\\ v_0^2 &=2g x_3\\ v_0 &= \sqrt{2gx_3} \ (\because v_0>0)\\ \end{aligned} Δx=v0t+12at2,Δx0t=v0t+12(g)t2xtx0=2gx3t12gt2xt=2gx3t12gt2 \begin{aligned} \Delta x &= v_0t+\frac{1}{2}at^2,\\ \Delta x_{0\to t} &= v_0t+\frac{1}{2}(-g)t^2\\ x_t-x_0 &= \sqrt{2gx_3}t-\frac{1}{2}gt^2\\ x_t &= \sqrt{2gx_3}t-\frac{1}{2}gt^2\\ \end{aligned} t=2g(x3±x3xt)(2-62-1)\therefore t = \sqrt{\frac{2}{g}}\left(\sqrt{x_3}\pm\sqrt{x_3-x_t}\right)\taag{2-62-1} (-Means UpDirection Time)\text{(-Means UpDirection Time)} (+Means DownDirection Time)\text{(+Means DownDirection Time)}
(a) t24=?t_{2\to4}=? Ans=t24=2t23=2(t3t2)=2{2g(x3x3x3)2g(x3x3x2)}=22g{x3(x3x3x2)}=22gx3x2=22gx3(x30.150[m])=22g0.150[m]=220.150[m]g=4015196133[s]0.3498084412322713[s]0.350[s] \begin{aligned} \Ans &= t_{2\to4} = 2t_{2\to3}=2(t_3-t_2)\\ &= 2\bra{\sqrt{\frac{2}{g}}\left(\sqrt{x_3}-\sqrt{x_3-x_3}\right)-\sqrt{\frac{2}{g}}\left(\sqrt{x_3}-\sqrt{x_3-x_2}\right)}\\ &= 2\sqrt{\frac{2}{g}}\bra{\sqrt{x_3}-\left(\sqrt{x_3}-\sqrt{x_3-x_2}\right)}\\ &= 2\sqrt{\frac{2}{g}}\sqrt{x_3-x_2}\\ &= 2\sqrt{\frac{2}{g}}\sqrt{x_3-(x_3-0.150\ut{m})}\\ &= 2\sqrt{\frac{2}{g}}\sqrt{0.150\ut{m}}\\ &= 2\sqrt{\frac{2\cdot0.150\ut{m}}{g}}\\ &=40 \sqrt{\frac{15}{196133}}\ut{s}\\ &\approx 0.3498084412322713\ut{s}\\ &\approx 0.350\ut{s} \end{aligned}
(b) t01+t56=?t_{0\to1}+t_{5\to6}=? Ans=2t01=2(t1t0)=2t1=22g(x3x3x1)=22g(0.780[m]0.780[m]0.150[m])=6(47+2546)g=6(47+2546)9.80665[s]=403(472546)196133[s]0.080792001405712[s]0.0808[s] \begin{aligned} \Ans &= 2t_{0\to1}=2(t_1-t_0)=2t_1\\ &= 2\sqrt{\frac{2}{g}}\left(\sqrt{x_3}-\sqrt{x_3-x_1}\right)\\ &= 2\sqrt{\frac{2}{g}}\left(\sqrt{0.780\ut{m}}-\sqrt{0.780\ut{m}-0.150\ut{m}}\right)\\ &=\sqrt{\frac{6}{(47+2 \sqrt{546})g}}\\ &=\sqrt{\frac{6}{(47+2 \sqrt{546})9.80665}}\ut{s}\\ &=40 \sqrt{\frac{3 \left(47-2\sqrt{546}\right)}{196133}}\ut{s}\\ &\approx 0.080792001405712\ut{s}\\ &\approx 0.0808\ut{s} \end{aligned} 점프 상단에서의 유지시간이 훨씬 길기 때문이다.