11판/2. 직선운동

2-49 할리데이 11판 솔루션 일반물리학

짱세디럭스 2024. 1. 17. 19:30
{H=45.0[m]t=2.00[s]h1=11.8[m]g=9.80665[m/s2] \begin{cases} H&=45.0\ut{m}\\ t&=2.00\ut{s}\\ h_1&=11.8\ut{m}\\ g&=9.80665\ut{m/s^2} \end{cases} S=vt12at2, S=vt-\frac{1}{2}at^2, h1H=v(2)12(g)(2)2(11.8)(45)=v(2)12(g)(2)2 \begin{aligned} h_1-H&=v(2)-\frac{1}{2}(-g)(2)^2\\ (11.8)-(45)&=v(2)-\frac{1}{2}(-g)(2)^2\\ \end{aligned} v=(835+g)=52813320000[m/s]=26.40665[m/s]26.4[m/s] \begin{aligned} \abs{v}&=\abs{-\(\frac{83}{5}+g\)}\\ &=\frac{528133}{20000}\ut{m/s}\\ &=26.40665\ut{m/s}\\ &\approx 26.4\ut{m/s}\\ \end{aligned}