10판/2. 직선운동

2-47 할리데이 10판 솔루션 일반물리학

짱세디럭스 2019. 7. 31. 01:43

{x0=98[m]x1=0[m]v0=14[m/s]a=g=9.80665[m/s2] \begin{cases} x_0 = 98\ut{m}\\ x_1 = 0\ut{m}\\ v_0 = 14\ut{m/s}\\ a = -g = -9.80665\ut{m/s^2}\\ \end{cases}
(a)v1=?\abs{v_1}=? 2aΔx=v2v02, 2a\Delta x = v^2-v_0^2, v2=2aΔx+v02 v^2 = 2a\Delta x +v_0^2 v1=2(g)(x0)+v02=2gx0+v02=2(9.80665[m/s2])(98[m])+(14[m/s])2=7502161332[m/s]46.02285736457484[m/s]46[m/s] \begin{aligned} \abs{v_1} &= \sqrt{2(-g)(-x_0) +v_0^2}\\ &= \sqrt{2gx_0+v_0^2}\\ &= \sqrt{2(9.80665\ut{m/s^2})(98\ut{m})+\(14\ut{m/s}\)^2}\\ &= \frac{7 }{50}\sqrt{\frac{216133}{2}}\ut{m/s}\\ &\approx 46.02285736457484\ut{m/s}\\ &\approx 46\ut{m/s} \end{aligned}
(b)t01=?t_{0\to1}=? Δx=v0t+12at2,Δx01=v0t01+12(g)t012x0=v0t112gt12 \begin{aligned} \Delta x &= v_0t+\frac{1}{2}at^2,\\ \Delta x_{0\to1} &= v_0t_{0\to1}+\frac{1}{2}(-g)t_{0\to1}^2\\ -x_0 &= v_0t_1-\frac{1}{2}gt_1^2\\ \end{aligned} t1=v0±2gx0+v02g=14[m/s]±2(9.80665[m/s2])98[m]+(14[m/s])29.80665[m/s2]=200(200+432266)28019[s](t1>0)6.120628080391859[s]6.1[s] \begin{aligned} t_1 &= \frac{v_0\pm\sqrt{2 g x_0+v_0^2}}{g}\\ &= \frac{14\ut{m/s}\pm\sqrt{2 (9.80665\ut{m/s^2}) 98\ut{m}+(14\ut{m/s})^2}}{9.80665\ut{m/s^2}}\\ &=\frac{200 \left(200+\sqrt{432266}\right)}{28019}\ut{s}(\because t_1>0)\\ &\approx 6.120628080391859\ut{s}\\ &\approx 6.1\ut{s} \end{aligned}