4-40 할리데이 10판 솔루션 일반물리학

$$\begin{cases} H &= 2.160\ut{m}\\ v_0 &= 15.00\ut{m/s}\\ \end{cases}$$ $$\begin{cases} \vec a &=-g\j\\ g &\approx 9.80665\ut{m/s^2}\\ \end{cases}$$ $$\begin{aligned} \Delta y&=v_{0y}t+\frac{1}{2}at^2\\ -H&=v_0t\sin\theta-\frac{1}{2}gt^2\\ -(2.160)&=(15.00)t\sin\theta-\frac{1}{2}gt^2\\ \end{aligned}$$ $$ \therefore t=\frac{3}{5 g} \left(\sqrt{12 g+625 \sin ^2\theta}+25 \sin \theta\right)$$ $$\begin{aligned} \Delta x &= v_x t\\ &=v_0\cos\theta \cdot t\\ &=(15.00)\cos\theta \cdot \frac{3}{5 g} \left(\sqrt{12 g+625 \sin ^2\theta}+25 \sin \theta\right)\\ &=\frac{9 }{g}\left(\sqrt{12 g+625 \sin ^2\theta}+25 \sin \theta\right)\cos \theta \end{aligned}$$
(a) $\theta_0=45.0\degree, \Delta x=?$ $$\begin{aligned} \Delta x &=\frac{9}{g}\left(\sqrt{12 g+625 \sin ^2\theta}+25 \sin \theta\right)\cos \theta\\ &= \frac{9 }{2 g}\left(\sqrt{24 g+625}+25\right)\\ &\approx 24.9313977643\ut{m}\\ &\approx 24.9\ut{m}\\ \end{aligned}$$
(b) $\theta_0=42.0\degree, \Delta x=?$ $$\begin{aligned} \Delta x &=\frac{9}{g}\left(\sqrt{12 g+625 \sin ^2\theta}+25 \sin \theta\right)\cos \theta\\ &\approx 25.0068642996\ut{m}\\ &\approx 25.0\ut{m}\\ \end{aligned}$$