11판/2. 직선운동

2-26 할리데이 11판 솔루션 일반물리학

짱세디럭스 2024. 1. 12. 18:31
$$ \begin{cases} H &= 145\ut{m}\\ h&=H-2R=144\ut{m}\\ g &= 9.80665\ut{m/s^2} \end{cases} $$ $$\ab{a}$$ $$ \begin{aligned} S&=v_0t+\frac{1}{2}at^2,\\ -h&=(0)t+\frac{1}{2}(-g)t^2\\ -144&=\frac{1}{2}(-g)t^2\\ \end{aligned} $$ $$ \begin{aligned} t&=\frac{2400}{\sqrt{196133}}\ut{s}\\ &\approx 5.41920906902145\ut{s}\\ &\approx 5.42\ut{s}\\ \end{aligned} $$ $$\ab{b}$$ $$ \begin{aligned} 2aS&=v^2-v_0^2,\\ 2(-g)(-h)&=v^2-(0)^2\\ 2g(144)&=v^2\\ v^2&=288g \end{aligned} $$ $$ \begin{aligned} v&=12\sqrt{2g}\\ &=\frac{3\sqrt{196133}}{25}\ut{m/s}\\ &\approx 53.1442866167192\ut{m/s}\\ &\approx 53.1\ut{m/s}\\ \end{aligned} $$ $$\ab{c}$$ $$ \begin{cases} a &= -25g\\ v &= 0 \end{cases} $$ $$ \begin{aligned} 2aS&=v^2-v_0^2,\\ 2(-25g)S&=(0)^2-290g\\ S&=\frac{29}{5}\ut{m}\\ &=5.80\ut{m}\\ \end{aligned} $$