11판/4. 2차원 운동과 3차원 운동

4-31 할리데이 11판 솔루션 일반물리학

짱세디럭스 2024. 2. 5. 17:46
{θ=70°d1=1.0[ft]d2=14[ft]h1=7.0[ft]h2=10[ft]g=9.80665[m/s2] \begin{cases} \theta&=70\degree\\ d_1 &= 1.0\ut{ft}\\ d_2 &= 14\ut{ft}\\ h_1 &= 7.0\ut{ft}\\ h_2 &= 10\ut{ft}\\ g&=9.80665\ut{m/s^2} \end{cases} 381[m]=1250[ft],381\ut{m}=1250\ut{ft}, g=1961336096[ft/s2]32.17404855643045[ft/s2]32[ft/s2] \begin{aligned} g&=\frac{196133}{6096}\ut{ft/s^2}\\ &\approx 32.17404855643045\ut{ft/s^2}\\ &\approx 32\ut{ft/s^2} \end{aligned} {Δx=d2d1=13[ft]Δy=h2h1=3.0[ft] \begin{cases} \Delta x &=d_2-d_1= 13\ut{ft}\\ \Delta y &=h_2-h_1=3.0\ut{ft}\\ \end{cases} {vx=v0cos70°vy=v0sin70° \begin{cases} v_x &= v_0\cos70\degree\\ v_y &= v_0\sin70\degree \end{cases} Δx=vxt,t=13[ft]v0cos70° \begin{aligned} \Delta x &= v_x t,\\ t&= \frac{13\ut{ft}}{v_0\cos70\degree} \end{aligned} S=v0t+12at2,Δy=vy0t+12(g)t23.0[ft]=(v0sin70°)t12gt2v0=13gcsc20°26cot20°626.65271277009674[ft/s]    27[ft/s]8.123746852325484[m/s]    8.1[m/s] \begin{aligned} S&=v_0t+\frac{1}{2}at^2,\\ \Delta y &=v_{y0}t+\frac{1}{2}(-g)t^2\\ 3.0\ut{ft} &=(v_0\sin70\degree)t-\frac{1}{2}gt^2\\ v_0&=\frac{13 \sqrt{g} \csc 20\degree}{\sqrt{26 \cot 20\degree-6}}\\ &\approx 26.65271277009674\ut{ft/s}\\ &~~~~\approx27\ut{ft/s}\\ &\approx 8.123746852325484\ut{m/s}\\ &~~~~\approx8.1\ut{m/s} \end{aligned}