10판/4. 2차원운동과 3차원운동

4-21 할리데이 10판 솔루션 일반물리학

짱세디럭스 2020. 7. 7. 15:36
{v0=6.75i^[m/s]Δr=5.00j^[cm]a=gj^g9.80665[m/s2]\begin{cases} \vec v_0 &= 6.75\i\ut{m/s}\\ \Delta \vec r &= -5.00\j\ut{cm}\\ \vec a &= -g\j\\ g &\approx 9.80665\ut{m/s^2}\\ \end{cases} Δx=? \Delta x =? Δy=vy0t+12ayt2,5.00×102=0t+12(g)t25.00×102=12gt2t=110g\begin{aligned} \Delta y &= v_{y0}t+\frac{1}{2}a_yt^2,\\ -5.00\times10^{-2} &= 0\cdot t+\frac{1}{2}(-g)t^2\\ 5.00\times10^{-2} &= \frac{1}{2}gt^2\\ t&= \frac{1}{\sqrt{10g}} \end{aligned} Δx=vxt=v0t=6.7510g=27410g274109.80665[m]0.681621742272[m]0.682[m]\begin{aligned} \Delta x &= v_x t = v_0 t\\ &= \frac{6.75}{\sqrt{10g}}\\ &= \frac{27}{4\sqrt{10g}}\\ &\approx \frac{27}{4\sqrt{10\cdot 9.80665}}\ut{m}\\ &\approx 0.681621742272\ut{m}\\ &\approx 0.682\ut{m}\\ \end{aligned}