2-65 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} d&=0.65\ut{m}\\ v_B&=\frac{1}{3}v_A\\ g&=9.80665\ut{m/s^2} \end{cases} $$ $$ \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} $$ $$ \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} $$