2-56 할리데이 10판 솔루션 일반물리학

$$\begin{cases} v_0 > 0 \\ v_B = \frac{v_A}{3}\\ \Delta y_{A \to B}=0.40\ut{m}\\ a=-g = -9.80665\ut{m/s^2}\\ \end{cases} $$ $$v_A =?$$ $$ 2a\Delta x = v^2-v_0^2,$$ $$ \begin{aligned} 2(-g)\Delta y_{A\to B} &= v_B^2-v_A^2\\ -2g\Delta y_{A\to B} &= \(\frac{v_A}{3}\)^2-v_A^2\\ &= -\frac{8}{9}v_A^2\\ \end{aligned} $$ $$ \begin{aligned} v_A^2 &= \frac{9}{4}g\Delta y_{A\to B}\\ v_A &= \sqrt{\frac{9}{4}g\Delta y_{A\to B}}\\ &= \sqrt{\frac{9}{4}(9.80665\ut{m/s^2})(0.40\ut{m})}\\ &=\frac{3}{200}\sqrt{\frac{196133}{5}}\ut{m/s}\\ &\approx 2.970855937267911\ut{m/s}\\ &\approx 2.97\ut{m/s} \end{aligned} $$