2-20 할리데이 11판 솔루션 일반물리학

$$\begin{cases} v&=18.0\ut{m/s}\\ H&=90.0\ut{m} \end{cases}$$ $$\ab{a}$$ $$\begin{aligned} S&=v_0t+\frac{1}{2}at^2,\\ -H&=(18)t+\frac{1}{2}(-g)t^2\\ \end{aligned}$$ $$\begin{aligned} t&=\frac{600 }{196133}\left(600+\sqrt{2321330}\right)\ut{s}\\ &\approx 6.496380983846016 \ut{s}\\ &\approx 6.50 \ut{s}\\ \end{aligned}$$ $$\ab{b}$$ $$\begin{aligned} 2aS&=v^2-v_0^2,\\ 2(-g)(-H)&=v^2-(18.0)^2\\ \end{aligned}$$ $$\begin{aligned} v&=\frac{3}{10}\sqrt{\frac{232133}{10}}\ut{m/s}\\ &\approx 45.70773457523354\ut{m/s}\\ &\approx 45.7\ut{m/s}\\ \end{aligned}$$