10판/4. 2차원운동과 3차원운동

4-37 할리데이 10판 솔루션 일반물리학

짱세디럭스 2020. 12. 1. 19:40
{y0=12[m]vx=2.50[m/s]\begin{cases} y_0 &= 12\ut{m}\\ v_x &= 2.50\ut{m/s}\\ \end{cases} {a=gj^g9.80665[m/s2]\begin{cases} \vec a &=-g\j\\ g &\approx 9.80665\ut{m/s^2}\\ \end{cases} {t0:Startt1=0.900[sec]t2:Finish\begin{cases} t_0 : \text{Start}\\ t_1=0.900\ut{sec}\\ t_2 : \text{Finish}\\ \end{cases}
(a) Δx1=?\Delta x_1=? Δx=vxt,=2.25[m]\begin{aligned} \Delta x &= v_xt,\\ &=2.25\ut{m} \end{aligned}
(b) y1=?y_1=? S=v0t+12at2,Δy=vy0t12gt2=12gt2\begin{aligned} S&=v_0t+\frac{1}{2}at^2,\\ \Delta y &= v_{y0}t-\frac{1}{2}gt^2\\ &=-\frac{1}{2}gt^2 \end{aligned} y1=y0+Δy,=y012gt2=1281g2008.02830675[m]8.03[m]\begin{aligned} y_1&=y_0+\Delta y,\\ &=y_0-\frac{1}{2}gt^2\\ &=12-\frac{81 g}{200}\\ &\approx 8.02830675\ut{m}\\ &\approx 8.03\ut{m}\\ \end{aligned}
(c) Δx2=?\Delta x_2=? S=v0t+12at2,y0=12gt2\begin{aligned} S&=v_0t+\frac{1}{2}at^2,\\ -y_0&=-\frac{1}{2}gt^2\\ \end{aligned} t=26g t=2\sqrt{\frac{6}{g}} Δx=vxt,=56g3.910977268492994[m]3.91[m]\begin{aligned} \Delta x &= v_x t,\\ &= 5\sqrt{\frac{6}{g}}\\ &\approx 3.910977268492994\ut{m}\\ &\approx 3.91\ut{m}\\ \end{aligned}