4-30 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} h&=6.20\ut{m}\\ v_0&=13.6\ut{m/s}\\ g&=9.80665\ut{m/s^2}\\ \end{cases} $$ $$ \begin{aligned} 2aS&=v^2-{v_0}^2,\\ 2(-g)(h)&=(0)^2-{v_{y0}}^2\\ v_{y0}&=\sqrt{2gh}\\ \end{aligned} $$ $$\ab{a}$$ $$ \begin{aligned} \theta &= \sin^{-1}\frac{v_{y0}}{v_0}\\ &= \sin^{-1}\frac{\sqrt{2gh}}{13.6}\\ &= \sin^{-1}\frac{\sqrt{2\cdot6.2g}}{13.6}\\ &\approx 0.9455767031562586\ut{rad}\\ &\approx 0.946\ut{rad}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} v&=v_0+at,\\ 0&=v_{y0}+(-g)t\\ t&=\frac{\sqrt{2gh}}{g}\\ &=\sqrt{\frac{2h}{g}}\\ \end{aligned} $$ $$ \begin{aligned} \Delta x &= v_x t\\ &=\sqrt{{v_0}^2-{v_{y0}}^2}t\\ &=\sqrt{{v_0}^2-2gh}\sqrt{\frac{2h}{g}}\\ &=\frac{2}{5} \sqrt{\frac{71672}{5g}-961}\\ &\approx 8.950548661002468\ut{m}\\ &\approx 8.95\ut{m}\\ \end{aligned} $$ $$\ab{c}$$ $$ \begin{aligned} \vec v_H&=v_x\i\\ v_H&=v_x\\ &= \sqrt{{v_0}^2-{v_{y0}}^2}\\ &= \sqrt{{v_0}^2-2gh}\\ &= \sqrt{{13.6}^2-2g(6.2)}\\ &\approx 7.959744970788952\ut{m/s}\\ &\approx 7.96\ut{m/s}\\ \end{aligned} $$