11판/2. 직선운동

2-6 할리데이 11판 솔루션 일반물리학

짱세디럭스 2023. 6. 23. 21:59
{H=100[m]v0=0 \begin{cases} H&=100\ut{m}\\ v_0&=0 \end{cases} (a)\ab{a} 2aS=v2v02,2aS=v^2-v_0^2, v=2aS+v02=2(g)(H)+(0)2=2gH=2(9.80665[m/s2])(100[m])=19613310[m/s]44.28690551393267[m/s]4×10[m/s] \begin{aligned} v&=\sqrt{2aS+v_0^2}\\ &=\sqrt{2(-g)(-H)+(0)^2}\\ &=\sqrt{2gH}\\ &=\sqrt{2(9.80665\ut{m/s^2})(100\ut{m})}\\ &=\frac{\sqrt{196133}}{10}\ut{m/s}\\ &\approx44.28690551393267\ut{m/s}\\ &\approx4\times10\ut{m/s}\\ \end{aligned} (b)\ab{b} S=v0t+12at2,S=v_0t+\frac{1}{2}at^2, t=2Sa=2(H)g=2Hg=2(100[m])(9.80665[m/s2])=2000196133[sec]4.516007557517876[sec]5[sec] \begin{aligned} t&=\sqrt{\frac{2S}{a}}\\ &=\sqrt{\frac{2(-H)}{-g}}\\ &=\sqrt{\frac{2H}{g}}\\ &=\sqrt{\frac{2(100\ut{m})}{(9.80665\ut{m/s^2})}}\\ &=\frac{2000}{\sqrt{196133}}\ut{sec}\\ &\approx4.516007557517876\ut{sec}\\ &\approx5\ut{sec}\\ \end{aligned} (c)\ab{c} 2aS=v2v02,2aS=v^2-v_0^2, v=2aS+v02=2(g)(H)+(0)2=2gH=2(9.80665[m/s2])(50[m])=1101961332[m/s]31.3155712066697[m/s]3×10[m/s] \begin{aligned} v&=\sqrt{2aS+v_0^2}\\ &=\sqrt{2(-g)(-H)+(0)^2}\\ &=\sqrt{2gH}\\ &=\sqrt{2(9.80665\ut{m/s^2})(50\ut{m})}\\ &=\frac{1}{10}\sqrt{\frac{196133}{2}}\ut{m/s}\\ &\approx31.3155712066697\ut{m/s}\\ &\approx3\times10\ut{m/s}\\ \end{aligned} (d)\ab{d} S=v0t+12at2,S=v_0t+\frac{1}{2}at^2, t=2Sa=2(H)g=2Hg=2(50[m])(9.80665[m/s2])=10002196133[sec]3.193299567810587[sec]3[sec] \begin{aligned} t&=\sqrt{\frac{2S}{a}}\\ &=\sqrt{\frac{2(-H)}{-g}}\\ &=\sqrt{\frac{2H}{g}}\\ &=\sqrt{\frac{2(50\ut{m})}{(9.80665\ut{m/s^2})}}\\ &=1000 \sqrt{\frac{2}{196133}}\ut{sec}\\ &\approx3.193299567810587\ut{sec}\\ &\approx3\ut{sec}\\ \end{aligned}