2-6 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} H&=100\ut{m}\\ v_0&=0 \end{cases} $$ $$\ab{a}$$ $$2aS=v^2-v_0^2,$$ $$ \begin{aligned} v&=\sqrt{2aS+v_0^2}\\ &=\sqrt{2(-g)(-H)+(0)^2}\\ &=\sqrt{2gH}\\ &=\sqrt{2(9.80665\ut{m/s^2})(100\ut{m})}\\ &=\frac{\sqrt{196133}}{10}\ut{m/s}\\ &\approx44.28690551393267\ut{m/s}\\ &\approx4\times10\ut{m/s}\\ \end{aligned} $$ $$\ab{b}$$ $$S=v_0t+\frac{1}{2}at^2,$$ $$ \begin{aligned} t&=\sqrt{\frac{2S}{a}}\\ &=\sqrt{\frac{2(-H)}{-g}}\\ &=\sqrt{\frac{2H}{g}}\\ &=\sqrt{\frac{2(100\ut{m})}{(9.80665\ut{m/s^2})}}\\ &=\frac{2000}{\sqrt{196133}}\ut{sec}\\ &\approx4.516007557517876\ut{sec}\\ &\approx5\ut{sec}\\ \end{aligned} $$ $$\ab{c}$$ $$2aS=v^2-v_0^2,$$ $$ \begin{aligned} v&=\sqrt{2aS+v_0^2}\\ &=\sqrt{2(-g)(-H)+(0)^2}\\ &=\sqrt{2gH}\\ &=\sqrt{2(9.80665\ut{m/s^2})(50\ut{m})}\\ &=\frac{1}{10}\sqrt{\frac{196133}{2}}\ut{m/s}\\ &\approx31.3155712066697\ut{m/s}\\ &\approx3\times10\ut{m/s}\\ \end{aligned} $$ $$\ab{d}$$ $$S=v_0t+\frac{1}{2}at^2,$$ $$ \begin{aligned} t&=\sqrt{\frac{2S}{a}}\\ &=\sqrt{\frac{2(-H)}{-g}}\\ &=\sqrt{\frac{2H}{g}}\\ &=\sqrt{\frac{2(50\ut{m})}{(9.80665\ut{m/s^2})}}\\ &=1000 \sqrt{\frac{2}{196133}}\ut{sec}\\ &\approx3.193299567810587\ut{sec}\\ &\approx3\ut{sec}\\ \end{aligned} $$