4-57 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} v_0&=40.0\ut{m/s}, \theta_0=30.0\degree\\ \Delta x &=150\ut{m}\\ g&=9.80665\ut{m/s^2} \end{cases} $$ $$ \begin{aligned} \Delta x&=v_x t,\\ 150&=40\cos30\degree t\\ t&=\frac{5\sqrt3}{2}\ut{s}\\ \end{aligned} $$ $$\ab{a}$$ $$ \begin{aligned} \Delta y&=v_0t+\frac{1}{2}at^2,\\ &=(40\sin30\degree)\(\frac{5\sqrt3}{2}\)+\frac{1}{2}(-g)\(\frac{5\sqrt3}{2}\)^2\\ &=50 \sqrt{3}-\frac{75 g}{8}\ut{m}\\ \end{aligned} $$ $$ \begin{aligned} h&=\abs{\Delta y}\\ &\approx 5.334803371556134\ut{m}\\ &\approx 5.33\ut{m}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} v&=v_0+at,\\ v_y&=40\sin30\degree-g\(\frac{5\sqrt3}{2}\)\\ &=20-\frac{5 \sqrt{3} g}{2}\\ \end{aligned} $$ $$ \begin{aligned} v&=\sqrt{{v_x}^2+{v_y}^2}\\ &=\sqrt{(20\sqrt{3})^2+\(20-\frac{5 \sqrt{3} g}{2}\)^2}\ut{m/s}\\ &\approx 41.28720260525459\ut{m/s}\\ &\approx 41.3\ut{m/s}\\ \end{aligned} $$