11판/4. 2차원 운동과 3차원 운동

4-9 할리데이 11판 솔루션 일반물리학

짱세디럭스 2024. 1. 27. 14:06

{va=23.0[m/s]vb=33.6[m/s]g=9.80665[m/s2] \begin{cases} v_a&=23.0\ut{m/s}\\ v_b&=33.6\ut{m/s}\\g&=9.80665\ut{m/s^2}\\ \end{cases} (a)\ab{a} va=vx,v_a = v_x, Δx=vxt=vat=23[m/s]5[s]=115[m] \begin{aligned} \Delta x &= v_x t\\ &=v_a t\\ &=23\ut{m/s}\cdot5\ut{s}\\ &=115\ut{m} \end{aligned} (b)\ab{b} vx2+vy2=v2,{v_x}^2+{v_y}^2=v^2, 2aS=v2v022(g)(H)=02vy022gH=vy02 \begin{aligned} 2aS&=v^2-{v_0}^2\\ 2(-g)(H)&=0^2-{v_{y0}}^2\\ 2gH&={v_{y0}}^2\\ \end{aligned} H=v2vx22g=33.622322g=1499950g=5999600196133[m]30.58944695691189[m]30.6[m] \begin{aligned} H&=\frac{v^2-{v_x}^2}{2g}\\ &=\frac{33.6^2-23^2}{2g}\\ &=\frac{14999}{50g}\\ &=\frac{5999600}{196133}\ut{m}\\ &\approx 30.58944695691189\ut{m}\\ &\approx 30.6\ut{m}\\ \end{aligned}