11판/2. 직선운동

2-65 할리데이 11판 솔루션 일반물리학

짱세디럭스 2024. 1. 19. 16:25
{d=0.65[m]vB=13vAg=9.80665[m/s2] \begin{cases} d&=0.65\ut{m}\\ v_B&=\frac{1}{3}v_A\\ g&=9.80665\ut{m/s^2} \end{cases} 2aS=v2v022(g)(d)=vB2vA22(g)(0.65)=(13vA)2vA2 \begin{aligned} 2aS&=v^2-{v_0}^2\\ 2(-g)(d)&={v_B}^2-{v_A}^2\\ 2(-g)(0.65)&=\(\frac{1}{3}v_A\)^2-{v_A}^2 \end{aligned} vA=3413g5=3400254972910[m/s]3.787113099050515[m/s]3.8[m/s] \begin{aligned} v_A&=\frac{3}{4} \sqrt{\frac{13g}{5}}\\ &=\frac{3 }{400}\sqrt{\frac{2549729}{10}}\ut{m/s}\\ &\approx 3.787113099050515\ut{m/s}\\ &\approx 3.8\ut{m/s}\\ \end{aligned}