4-37 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} v_0&=35.6\ut{m/s}\\ R&=26.4\ut{m}\\ g&=9.80665\ut{m/s^2} \end{cases} $$ $$ \begin{aligned} S&=v_0t+\frac{1}{2}at^2,\\ \Delta y &=v_{y0}t+\frac{1}{2}(-g)t^2\\ 0&=v_{y0}t-\frac{1}{2}gt^2\\ &=v_{y0}-\frac{1}{2}gt\\ \end{aligned} $$ $$t=\frac{2v_{y0}}{g}=\frac{2v_0\sin\theta}{g}$$ $$ \begin{aligned} \Delta x &= v_x t\\ R &= (v_0\cos\theta)\(\frac{2v_0\sin\theta}{g}\)\\ &=\frac{{v_0}^2\sin(2\theta)}{g} \end{aligned} $$ $$ \begin{aligned} \sin(2\theta)&=\frac{gR}{{v_0}^2},\\ \end{aligned} $$ $$ \begin{cases} 2\theta_1 &= \sin^{-1}\cfrac{gR}{{v_0}^2}\\ 2\theta_2 &= \pi - 2\theta_1 \end{cases} $$ $$ \begin{cases} \theta_1 &= \cfrac{1}{2}\sin^{-1}\cfrac{gR}{{v_0}^2}\\ \theta_2 &= \cfrac{\pi}{2} - \cfrac{1}{2}\sin^{-1}\cfrac{gR}{{v_0}^2} \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} \theta_1 &= \frac{1}{2}\sin^{-1}\frac{gR}{{v_0}^2}\\ &=\frac{1}{2}\sin^{-1}\(\frac{6472389}{31684000}\)\\ &\approx 0.1028637717788844\ut{rad}\\ &\approx0.103\ut{rad} \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} \theta_2 &= \frac{\pi}{2} - \frac{1}{2}\sin^{-1}\frac{gR}{{v_0}^2}\\ &= \frac{\pi}{2} - \frac{1}{2}\sin^{-1}\(\frac{6472389}{31684000}\)\\ &\approx 1.467932555016012\ut{rad}\\ &\approx 1.47\ut{rad} \end{aligned} $$