2-16 할리데이 11판 솔루션 일반물리학

$$\begin{cases} v_A&=v\j\\ v_B&=\frac{1}{2}v\j\\ y_B&=y_A+4\ut{m}\\ g&=9.80665\ut{m/s^2} \end{cases}$$ $$\ab{a}$$ $$2aS=v^2-v_0^2,$$ $$\begin{aligned} 2aS&=v^2-v_0^2\\ 2(-g)(4\ut{m})&=\(\frac{1}{2}v\)^2-(v)^2\\ \end{aligned}$$ $$\begin{aligned} v&=\sqrt{\frac{32g}{3}}\\ &=\frac{1}{25}\sqrt{\frac{196133}{3}}\ut{m/s}\\ &\approx 10.22762272801782\ut{m/s}\\ &\approx 1\times10\ut{m/s}\\ \end{aligned}$$