11판/4. 2차원 운동과 3차원 운동

4-60 할리데이 11판 솔루션 일반물리학

짱세디럭스 2024. 2. 13. 20:43
{v0=12.0[m/s],θ0=15.0°t1=1.51[s] \begin{cases} v_0&=12.0\ut{m/s},\theta_0=-15.0\degree\\ t_1 &= 1.51\ut{s} \end{cases} (a)\ab{a} vx=12cos(15°)=32(1+3)[m/s] \begin{aligned} v_x &= 12\cos(-15\degree)\\ &=3 \sqrt{2} \(1+\sqrt{3}\)\ut{m/s}\\ \end{aligned} Δx=vxt=32(1+3)1.51=453(1+3)502[m]17.50257597235792[m]17.5[m] \begin{aligned} \Delta x &= v_x t\\ &=3 \sqrt{2} \(1+\sqrt{3}\)\cdot 1.51\\ &=\frac{453 \(1+\sqrt{3}\)}{50 \sqrt{2}}\ut{m}\\ &\approx 17.50257597235792\ut{m}\\ &\approx 17.5\ut{m}\\ \end{aligned} (b)\ab{b} vy0=12sin(15°)=32(31)[m/s] \begin{aligned} v_{y0}&=12\sin(-15\degree)\\ &=-3\sqrt{2}\(\sqrt{3}-1\)\ut{m/s}\\ \end{aligned} S=v0t+12at2,Δy=vy0t+12(g)t2=32(31)(1.51)12g(1.51)215.86987242975767[m]15.9[m] \begin{aligned} S&=v_0t+\frac{1}{2}at^2,\\ \Delta y&=v_{y0}t+\frac{1}{2}(-g)t^2\\ &=-3\sqrt{2}\(\sqrt{3}-1\)(1.51)-\frac{1}{2}g(1.51)^2\\ &\approx -15.86987242975767\ut{m}\\ &\approx -15.9\ut{m}\\ \end{aligned}