4-60 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} v_0&=12.0\ut{m/s},\theta_0=-15.0\degree\\ t_1 &= 1.51\ut{s} \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} v_x &= 12\cos(-15\degree)\\ &=3 \sqrt{2} \(1+\sqrt{3}\)\ut{m/s}\\ \end{aligned} $$ $$ \begin{aligned} \Delta x &= v_x t\\ &=3 \sqrt{2} \(1+\sqrt{3}\)\cdot 1.51\\ &=\frac{453 \(1+\sqrt{3}\)}{50 \sqrt{2}}\ut{m}\\ &\approx 17.50257597235792\ut{m}\\ &\approx 17.5\ut{m}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} v_{y0}&=12\sin(-15\degree)\\ &=-3\sqrt{2}\(\sqrt{3}-1\)\ut{m/s}\\ \end{aligned} $$ $$ \begin{aligned} S&=v_0t+\frac{1}{2}at^2,\\ \Delta y&=v_{y0}t+\frac{1}{2}(-g)t^2\\ &=-3\sqrt{2}\(\sqrt{3}-1\)(1.51)-\frac{1}{2}g(1.51)^2\\ &\approx -15.86987242975767\ut{m}\\ &\approx -15.9\ut{m}\\ \end{aligned} $$