8-15 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} m&=425\ut{g}\\ L&=2.4\ut{m}\\ \theta_i&=30.0\degree\\ v_i&=0\\ g&=9.80665\ut{m/s^2}\\ \end{cases} $$ $$ \put \begin{cases} \KE : \text{Kinetic Energy}\\ \GE : \text{Gravitational Potential Energy}\\ \end{cases} $$ $$\Sigma E_i=\Sigma E_f,$$ $$ \begin{aligned} \KE_i+\GE_i&=\KE_f+\GE_f\\ mgy_i&=\frac{1}{2}m{v_f}^2+mgy_f\\ 2gy_i&={v_f}^2+2gy_f\\ \end{aligned} $$ $$y=L-L\cos\theta=L(1-\cos\theta),$$ $$ v_f=\sqrt{2g(y_i-y_f)},$$ $$ \begin{aligned} v_f&=\sqrt{2g\bra{(L-L\cos\theta_i)-(L-L\cos\theta_f)}}\\ &=\sqrt{2gL(\cos\theta_f-\cos\theta_i)}\\ \end{aligned} $$ $$\therefore v(\theta)=2 \sqrt{\frac{3g}{5}(2 \cos \theta-\sqrt{3} )}$$ $$\ab{a}$$ $$ \begin{aligned} v(20.0\degree)&=2 \sqrt{\frac{3g}{5}(2 \cos 20.0\degree-\sqrt{3} )}\\ &\approx 1.8621646933920286\ut{m/s}\\ &\approx 1.9\ut{m/s}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} \max(v)&=v(0\degree)\\ &=2 \sqrt{\frac{3g}{5}(2-\sqrt{3} )}\\ &\approx 2.5112629243253703\ut{m/s}\\ &\approx 2.5\ut{m/s}\\ \end{aligned} $$ $$\ab{c}$$ $$v(\phi)=\frac{1}{3}\max(v)$$ $$ 2 \sqrt{\frac{3g}{5}(2 \cos \phi-\sqrt{3} )}=\frac{1}{3}\cdot 2 \sqrt{\frac{3g}{5}(2-\sqrt{3} )} $$ $$\cos \phi=\frac{1+4\sqrt3}{9}$$ $$ \begin{aligned} \phi&=0.49301171100758445\ut{rad}\\ &=0.49\ut{rad}\\ \end{aligned} $$