2-59 할리데이 10판 솔루션 일반물리학

$$\begin{cases} d = \text{Water Drop Interval Time}\\ A,B,C,D = \text{1,2,3,4st Water}\\ t_N = \text{Nst Water Start}\\ t_4 = \text{1st Water Landing}\\ \end{cases}$$ $$\begin{cases} h=x_{A1}=x_{B2}=x_{C3}=x_{D4}=0.2\ut{m}\\ x_{A4}=0\\ t_1 = 0\\ t_2 = d\\ t_3 = d+t_2=2d\\ t_4 = d+t_3=3d\\ v_0=v_{A1}=v_{B2}=v_{C3}=v_{D4}=0\\ a=-g = -9.80665\ut{m/s^2}\\ \end{cases}$$ $$\begin{aligned} \Delta x &= v_0t+\frac{1}{2}at^2,\\ \Delta x_{A1\to4} &= v_{A1}t_{1\to4}+\frac{1}{2}(-g)t_{1\to4}^2\\ x_{A4}-x_{A1} &= v_{A1}(t_4-t_1)-\frac{1}{2}g(t_4-t_1)^2\\ 0-h &= 0(3d-0)-\frac{1}{2}g(3d-0)^2\\ h &= \frac{9}{2}gd^2\\ \end{aligned}$$ $$d^2=\frac{2h}{9g}\taag{2-59-1}$$ $$\begin{aligned} \Delta x &= v_0t+\frac{1}{2}at^2,\\ \Delta x_{1\to t} &= v_0t_{1\to t}+\frac{1}{2}(-g)t_{1\to t}^2\\ x_t-x_1 &= v_0(t-t_1)-\frac{1}{2}g(t-t_1)^2\\ x(t)-h &= -\frac{1}{2}gt^2\\ x(t) &= h-\frac{1}{2}gt^2\taag{2-59-2}\\ \end{aligned}$$
(a)$x_{B2\to4}=?$ $$\begin{aligned} \Ans &=x_{B4}-x_{B2}\\ &=x(t_4-t_2)-h\\ &=x(3d-d)-h\\ &=h-\frac{1}{2}g(2d)^2-h\reef{2-59-2}\\ &=-2gd^2\\ &=-2g\(\frac{2h}{9g}\)\reef{2-59-1}\\ &=-\frac{4}{9}h\\ &=-\frac{4}{9}(0.2\ut{m})\\ &=-\frac{4}{45}\ut{m}\\ &\approx 0.08888888888888889\ut{m}\\ &\approx 9\ut{cm}\\ \end{aligned}$$
(b)$x_{C3\to4}=?$ $$\begin{aligned} \Ans &=x_{C4}-x_{C3}\\ &=x(t_4-t_3)-h\\ &=x(3d-2d)-h\\ &=h-\frac{1}{2}g(d)^2-h\reef{2-59-2}\\ &=-\frac{1}{2}gd^2\\ &=-\frac{1}{2}g\(\frac{2h}{9g}\)\reef{2-59-1}\\ &=-\frac{1}{9}h\\ &=-\frac{1}{9}(0.2\ut{m})\\ &=-\frac{1}{45}\ut{m}\\ &\approx 0.02222222222222222\ut{m}\\ &\approx 2\ut{cm}\\ \end{aligned}$$