4-7 할리데이 11판 솔루션 일반물리학

$$ \begin{cases} \theta_0&=60.0\degree\\ h&=2.60\ut{km}\\ d&=9.40\ut{km}\\ g&=9.80665\ut{m/s^2}\\ \end{cases} $$ $$ \begin{aligned} \Delta x &= v_x t,\\ d&= v_0\cos\theta t \end{aligned} $$ $$ \begin{aligned} S&=v_0t+\frac{1}{2}at^2,\\ \Delta y &=v_{y0}t+\frac{1}{2}(-g)t^2\\ -h &=v_0\sin\theta t-\frac{1}{2}gt^2\\ \end{aligned} $$ $$ \begin{cases} d&= v_0\cos\theta t\\ -h &=v_0\sin\theta t-\frac{1}{2}gt^2\\ \end{cases} $$ $$ \begin{cases} 9.4\times10^3&= v_0\cos(60\degree) t\\ -2.6\times10^3 &=v_0\sin(60\degree) t-\frac{1}{2}gt^2\\ \end{cases} $$ $$ \begin{cases} v_0&=\cfrac{940 \sqrt{g}}{\sqrt{13+47 \sqrt{3}}}\\ t&=\cfrac{20 \sqrt{13+47\sqrt{3}}}{\sqrt{g}} \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} v_0&=\frac{940 \sqrt{g}}{\sqrt{13+47 \sqrt{3}}}\\ &= \frac{47}{5} \sqrt{\frac{196133}{26+94 \sqrt{3}}}\ut{m/s}\\ &\approx 302.9615406215586\ut{m/s}\\ &\approx 303\ut{m/s}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} t&=\frac{20 \sqrt{13+47\sqrt{3}}}{\sqrt{g}}\\ &=2000 \sqrt{\frac{2 \left(13+47 \sqrt{3}\right)}{196133}}\ut{s}\\ &\approx 62.0540810606843\ut{s}\\ &\approx 62.1\ut{s}\\ \end{aligned} $$ $$\ab{c}$$ $$\title{More Big Answer}$$