2-18 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} \text{A : in Air}\\ \text{B : in Water}\\ \end{cases} $$ $$ \begin{cases} H_A&=7.00\ut{m}\\ t_B&=4.00\ut{s}\\ g&=9.80665\ut{m/s^2} \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} 2aS&=v^2-v_0^2,\\ 2(-g)(-H)&=v^2-(0)^2 \end{aligned} $$ $$ \begin{aligned} v_B&=\sqrt{2gH}\\ &=\sqrt{14g} \end{aligned} $$ $$ \begin{aligned} H_B&=v_B\cdot t\\ &=\(\sqrt{14g}\)\cdot (4\ut{s})\\ &=\frac{7}{25}\sqrt{28019}\ut{m}\\ &\approx46.8688553305924\ut{m}\\ &\approx46.9\ut{m}\\ \end{aligned} $$ $$\ab{b}$$ $$ \Ans = \bar v = \frac{\Sigma S}{\Sigma t},$$ $$S=v_0t+\frac{1}{2}at^2,$$ $$ \begin{aligned} -H&=(0)t+\frac{1}{2}(-g)t^2\\ H&=\frac{1}{2}gt^2 \end{aligned} $$ $$ \begin{aligned} t_A&=\sqrt{\frac{2H}{g}}\\ &=\sqrt{\frac{2(7.00\ut{m})}{g}}\\ &=\frac{200}{\sqrt{28019}}\ut{s} \end{aligned} $$ $$ \begin{aligned} \Sigma S&=H_A+H_B\\ &=7.00\ut{m}+\frac{7}{25}\sqrt{28019}\ut{m}\\ \Sigma t&=t_A+t_B\\ &=\frac{200}{\sqrt{28019}}\ut{s}+4.00\ut{s}\\ \bar v &= \frac{\Sigma S}{\Sigma t}\\ &=\frac{7.00\ut{m}+\frac{7}{25}\sqrt{28019}\ut{m}}{\frac{200}{\sqrt{28019}}\ut{s}+4.00\ut{s}}\\ &=\frac{7 }{2551900}\left(26769 \sqrt{28019}-700475\right)\ut{m/s}\\ &\approx10.3697185268293\ut{m/s}\\ &\approx10.4\ut{m/s}\\ \end{aligned} $$ $$\ab{c}$$ $$\title{Down}$$ $$\ab{d}$$ $$ \begin{cases} H&=7.00\ut{m}+\frac{7}{25}\sqrt{28019}\\ t&=4.00\ut{s}\\ g&=9.80665\ut{m/s^2} \end{cases} $$ $$S=v_0t+\frac{1}{2}at^2,$$ $$ \begin{aligned} -H&=v_0(4)+\frac{1}{2}(-g)(4)^2\\ \end{aligned} $$ $$ \begin{aligned} v_0&=\frac{7 }{10000}\left(25519-100 \sqrt{28019}\right)\ut{m/s}\\ &\approx6.146086167351901\ut{m/s}\\ &\approx6.15\ut{m/s}\\ \end{aligned} $$ $$\ab{e}$$ $$\title{Up}$$