10판/4. 2차원운동과 3차원운동

4-31 할리데이 10판 솔루션 일반물리학

짱세디럭스 2020. 8. 11. 19:44
{rQ=(32,3.5)[m]rP=(0,1.0)[m]v0=(18[m/s],40°)a=gj^g9.80665[m/s2]\begin{cases} \vec r_Q &= (32,3.5)\ut{m}\\ \vec r_P &= (0,1.0)\ut{m}\\ \vec v_0 &= (18\ut{m/s},40\degree)\\ \vec a &=-g\j\\ g &\approx 9.80665\ut{m/s^2}\\ \end{cases} [Over Q Point Possible?]\title{Over Q Point Possible?} v0=18cos40°i^+18sin40°j^,v_0 = 18\cos40\degree\i+18\sin40\degree\j, t=Δxvx=3218cos40°=169sec40°\begin{aligned} t &= \frac{\Delta x}{v_x}\\ &= \frac{32}{18\cos40\degree}\\ &= \frac{16}{9} \sec40\degree\\ \end{aligned} Δy=vy0t+12ayt2,=(18sin40°)(169sec40°)12g(169sec40°)2=32tan40°12881gsec240°0.443055522227[m]<2.5[m]\begin{aligned} \Delta y &= v_{y0}t+\frac{1}{2}a_yt^2,\\ &= \(18\sin40\degree\) \(\frac{16}{9} \sec40\degree\)-\frac{1}{2}g\(\frac{16}{9} \sec40\degree\)^2\\ &=32 \tan 40\degree-\frac{128}{81} g \sec ^2 40\degree\\ &\approx0.443055522227\ut{m} < 2.5\ut{m} \end{aligned} [Over Q Point Impossible]\title{Over Q Point Impossible}