4-28 할리데이 11판 솔루션 일반물리학

$$\begin{cases} v_0&=16.0\ut{m/s}, \theta_0\\g&=9.80665\ut{m/s^2} \end{cases}$$ $$\ab{a,b}$$ $$\begin{aligned} \Delta x=v_xt\\ =v_0\cos\theta_0t \end{aligned}$$ $$\begin{aligned} S=v_0t+\frac{1}{2}at^2,\\ \Delta y=v_{y0}t+\frac{1}{2}(-g)t^2\\ =(v_0\sin\theta_0)t-\frac{1}{2}gt^2\\ \end{aligned}$$ $$\Delta y=(v_0\sin\theta_0)t-\frac{1}{2}gt^2\ge 0,$$ $$\therefore 0\le t \le \frac{32 \sin \theta_0}{g}$$ $$\begin{aligned} r(\theta_0,t)=\sqrt{\(v_0\cos\theta_0t\)^2+\bra{(v_0\sin\theta_0)t-\frac{1}{2}gt^2}^2}\\ =\frac{1}{2} \sqrt{t^2 \(g^2 t^2-4 g t v_0 \sin\theta_0+4 {v_0}^2\)}\\ =\frac{1}{2} \sqrt{t^2 \(g^2 t^2-64 g t \sin\theta_0+1024\)}\\ \end{aligned}$$ $$\ab{a}$$ $$\begin{aligned} r(40\degree,t)=\frac{1}{2} \sqrt{t^2 \(g^2 t^2-64 g t \sin40\degree+1024\)}\\ \approx \sqrt{145.324 t^2+10.2846 t}\\ t \le \frac{32\sin40\degree}{g}\approx 2.09748\ut{s}\\ \end{aligned}$$ $$\ab{b}$$ $$\begin{aligned} r(80\degree,t)=\frac{1}{2} \sqrt{t^2 \(g^2 t^2-64 g t \sin80\degree+1024\)}\\ \approx \sqrt{2.81602 t^2+15.7569 t}\\ t \le \frac{32\sin80\degree}{g}\approx 3.21352\ut{s}\\ \end{aligned}$$ 

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$$\begin{aligned} \dxt{r}&=\dt\(\frac{1}{2} \sqrt{t^2 \(g^2 t^2-64 g t \sin\theta_0+1024\)}\)\\ &=\frac{t \(g^2 t^2-48 g t \sin\theta_0+512\)}{\sqrt{t^2 \(g^2 t^2-64 g t \sin\theta_0+1024\)}}\\ \end{aligned}$$ $$\ab{c,d}$$ $$\begin{aligned} \dt r(40\degree,t)&=\frac{t \(g^2 t^2-48 g t \sin40\degree+512\)}{\sqrt{t^2 \(g^2 t^2-64 g t \sin40\degree+1024\)}}\gt0\\ &\(t\le\frac{32\sin40\degree}{g}\)\\ \end{aligned}$$ $$\begin{aligned} \therefore r(40\degree,t)_{\max}&=r\(40\degree,\frac{32\sin40\degree}{g}\)\\ \end{aligned}$$ $$\ab{c}$$ $$\begin{aligned} t_{40}&=\frac{32\sin40\degree}{g}\\ &\approx 2.097475030715816\ut{s}\\ &\approx 2.10\ut{s}\\ \end{aligned}$$ $$\ab{d}$$ $$\begin{aligned} r(40\degree,t)&=\frac{1}{2} \sqrt{t^2 \(g^2 t^2-64 g t \sin40\degree+1024\)},\\ \Ans&=r\(40\degree,\frac{32\sin40\degree}{g}\)\\ &\approx 25.70814546977054\ut{m}\\ &\approx 25.7\ut{m}\\ \end{aligned}$$ $$\ab{e,f}$$ $$\begin{aligned} \vec r(\theta_0,t) &= (v_0\cos\theta_0t)\i+\(v_0\sin\theta_0t-\frac{1}{2}gt^2\)\j\\ \vec r_{\max}&=\vec r\(40\degree,\frac{32\sin40\degree}{g}\)\\ &=\frac{256 \cos10\degree}{g}\i\\ \end{aligned}$$ $$\ab{e}$$ $$\begin{aligned} \Ans&=r_x\(40\degree,\frac{32\sin40\degree}{g}\)\\ &=\frac{256 \cos10\degree}{g}\\ &\approx 25.70814546977054\ut{m}\\ &\approx 25.7\ut{m} \end{aligned}$$ $$\ab{f}$$ $$\begin{aligned} \Ans&=r_y\(40\degree,\frac{32\sin40\degree}{g}\)\\ &=0\\ \end{aligned}$$ $$\ab{g,h}$$ $$\begin{aligned} \dt r(80\degree,t)&=\frac{t \(g^2 t^2-48 g t \sin80\degree+512\)}{\sqrt{t^2 \(g^2 t^2-64 g t \sin80\degree+1024\)}}=0\\ &\(t\le\frac{32\sin80\degree}{g}\)\\ \end{aligned}$$ $$\begin{aligned} t&=\frac{4 }{g}\(6 \cos 10\degree\pm\sqrt{18 \cos 20\degree-14}\)\\ t_{\max}&= \frac{4 }{g}\(6 \cos 10\degree-\sqrt{18 \cos 20\degree-14}\) \end{aligned}$$ $$\ab{g}$$ $$\begin{aligned} t_{\max}&= \frac{4 }{g}\(6 \cos 10\degree-\sqrt{18 \cos 20\degree-14}\)\\ &\approx 1.713802555341173\ut{s}\\ &\approx 1.71\ut{s}\\ \end{aligned}$$ $$\ab{h}$$ $$\begin{aligned} r(80\degree,t)&=\frac{1}{2} \sqrt{t^2 \(g^2 t^2-64 g t \sin80\degree+1024\)},\\ \Ans&=r\(80\degree,t_{\max}\)\\ &\approx 13.47213163396955\ut{m}\\ &\approx 13.5\ut{m}\\ \end{aligned}$$ $$\ab{i,j}$$ $$\begin{aligned} \vec r(\theta_0,t) &= (v_0\cos\theta_0t)\i+\(v_0\sin\theta_0t-\frac{1}{2}gt^2\)\j\\ \vec r_{\max}=&\vec r\(80\degree,t_{\max}\)\\ =&-\frac{64 \sin10\degree \(\sqrt{2 (9 \cos20\degree-7)}-6 \cos10\degree\)}{g}\i\\&+\frac{32 \(-3 \cos20\degree+\cos10\degree\sqrt{2 (9 \cos20\degree-7)}+5\)}{g}\j \end{aligned}$$ $$\ab{i}$$ $$\begin{aligned} \Ans&=r_x\(80\degree,t_{\max}\)\\ &=-\frac{64 \sin10\degree \(\sqrt{2 (9 \cos20\degree-7)}-6 \cos10\degree\)}{g}\\ &\approx 4.761579049854769\ut{m}\\ &\approx 4.76\ut{m} \end{aligned}$$ $$\ab{f}$$ $$\begin{aligned} \Ans&=r_y\(80\degree,t_{\max}\)\\ &=\frac{32 \(-3 \cos20\degree+\cos10\degree\sqrt{2 (9 \cos20\degree-7)}+5\)}{g}\\ &\approx 12.602606703178\ut{m}\\ &\approx 12.6\ut{m} \end{aligned}$$