8-33 할리데이 11판 솔루션 일반물리학

$$ \put \begin{cases} \KE : \text{Kinetic Energy}\\ \GE : \text{Gravitational Potential Energy}\\ \end{cases} $$ $$ \put \begin{cases} 0 : \text{Start}\\ 1 : v_y=65\ut{m/s}\\ 2 : \text{Highest Point} \end{cases} $$ $$ \begin{cases} m&=0.650\ut{kg}\\ \KE_0 &= 1700\ut{J}\\ \Delta h_{0\rarr2}&=+180\ut{m}\\ g&=9.80665\ut{m/s^2}\\ \end{cases} $$ $$\ab{a}$$ $$\vec v_2 = v_x\i+0\j,$$ $$ \therefore v_2=v_x$$ $$\Sigma E_0=\Sigma E_2,$$ $$ \begin{aligned} \KE_0+\GE_0&=\KE_2+\GE_2\\ \KE_2&=\KE_0-\Delta \GE\\ \frac{1}{2}m{v_2}^2&=\KE_0-\Delta (mgh)\\ \end{aligned} $$ $$ \begin{aligned} v_x&=\sqrt{\frac{2(\KE_0-mg\Delta h)}{m}}\\ &=2 \sqrt{\frac{17000}{13}-90 g}\\ &\approx 41.235606346569355\ut{m/s}\\ &\approx 41\ut{m/s}\\ \end{aligned} $$ $$\ab{b}$$ $$2aS=v^2-{v_0}^2,$$ $$2(-g)(\Delta h)=0^2-{v_{0y}}^2$$ $$ \begin{aligned} v_{0y}&=\sqrt{2g\Delta h}\\ &=6\sqrt{10g}\\ &\approx 59.41711874535822\ut{m/s}\\ &\approx 59\ut{m/s} \end{aligned} $$ $$\ab{c}$$ $$2aS=v^2-{v_0}^2,$$ $$2(-g)S={v_1}^2-{v_0}^2$$ $$ \begin{aligned} S&=\frac{{v_1}^2-{v_0}^2}{-2g}\\ &=180-\frac{4225}{2g}\\ &\approx -35.41504999158734\ut{m}\\ &\approx -35\ut{m}\\ \end{aligned} $$