2-34 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} h_B&=h_A+15\ut{m}\\ t_{A0} &= 0\ut{s}\\ t_{B0} &= 1\ut{s}\\ \Delta y_3 &=0\\ g&=9.80665\ut{m/s^2} \end{cases} $$ $$S=v_0t+\frac{1}{2}at^2,$$ $$ \begin{cases} S_A&=v_{A0}t_A+\frac{1}{2}a_A{t_A}^2\\ S_B&=v_{B0}t_B+\frac{1}{2}a_B{t_B}^2\\ \end{cases} $$ $$ \begin{cases} -\Delta y_A&=(0)t_A+\frac{1}{2}(-g)(t_B+1)^2\\ -\Delta y_B&=v_{B0}{t_B}+\frac{1}{2}(-g){t_B}^2\\ \end{cases} $$ $$ \begin{cases} -\Delta y_A&=-\frac{1}{2}g(t_B+1)^2\\ -\Delta y_B&=v_{B0}{t_B}-\frac{1}{2}g{t_B}^2\\ \end{cases} $$ $$ \begin{aligned} &\Delta y_A-\Delta y_B=15\\ &=\bra{\frac{1}{2}g(t_B+1)^2}+\bra{v_{B0}{t_B}-\frac{1}{2}g{t_B}^2}\\ \end{aligned} $$ $$ \begin{aligned} v_{B0}&=\frac{-2 g \Delta t_B-g+30}{2 \Delta t_B}\\ &= \frac{1}{4} (30-5 g)\\ &=-\frac{76133}{16000}\ut{m/s}\\ &=-4.7583125\ut{m/s}\\ \abs{v_{B0}}&\approx4.8\ut{m/s}\\ \end{aligned} $$