10판/4. 2차원운동과 3차원운동

4-26 할리데이 10판 솔루션 일반물리학

짱세디럭스 2020. 7. 27. 20:49
{v0=(18.0[m/s],40.0°)a=gj^g9.80665[m/s2]\begin{cases} \vec v_0 &= (18.0\ut{m/s},40.0\degree)\\ \vec a &= -g\j\\ g &\approx 9.80665\ut{m/s^2}\\ \end{cases} v0=(18.0[m/s],40.0°)=18cos40°i^+18sin40°j^\begin{aligned} \vec v_0 &= (18.0\ut{m/s},40.0\degree)\\ &=18\cos40\degree\i+ 18\sin40\degree\j\\ \end{aligned} Δy=vy0t+12ayt2,=18sin40°t+12(g)t2=18sin40°t12gt2\begin{aligned} \Delta y &= v_{y0}t+\frac{1}{2}a_yt^2,\\ &= 18\sin40\degree t+\frac{1}{2}(-g)t^2\\ &= 18\sin40\degree t-\frac{1}{2}gt^2\\ \end{aligned} r=(18cos40°t)i^+(18sin40°t12gt2)j^ \vec r = \(18\cos40\degree t\)\i+\(18\sin40\degree t-\frac{1}{2}gt^2\)\j
(a) Δxt=1.10=?\Delta x_{t=1.10}=? Δx=18cos40°t=18cos40°1.10=995cos40°[m]15.1676799738[m]15.2[m]\begin{aligned} \Delta x &= 18\cos40\degree t\\ &= 18\cos40\degree \cdot 1.10\\ &=\frac{99}{5} \cos 40\degree\ut{m}\\ &\approx 15.1676799738\ut{m}\\ &\approx 15.2\ut{m} \end{aligned}
(b) Δyt=1.10=?\Delta y_{t=1.10}=? Δy=18sin40°t12gt2=18sin40°1.1012g(1.10)2=995sin40°121200g6.79417142179[m]6.79[m]\begin{aligned} \Delta y &= 18\sin40\degree t-\frac{1}{2}gt^2\\ &= 18\sin40\degree \cdot 1.10-\frac{1}{2}g(1.10)^2\\ &=\frac{99}{5} \sin40 \degree-\frac{121}{200}g\\ &\approx6.79417142179\ut{m}\\ &\approx6.79\ut{m}\\ \end{aligned}
(c) Δxt=1.80=?\Delta x_{t=1.80}=? Δx=18cos40°t=18cos40°1.80=1625cos40°[m]24.8198399571[m]24.8[m]\begin{aligned} \Delta x &= 18\cos40\degree t\\ &= 18\cos40\degree \cdot 1.80\\ &=\frac{162}{5} \cos40\degree\ut{m}\\ &\approx 24.8198399571\ut{m}\\ &\approx 24.8\ut{m}\\ \end{aligned}
(d) Δyt=1.80=?\Delta y_{t=1.80}=? Δy=18sin40°t12gt2=18sin40°1.8012g(1.80)2=1625sin40°8150g4.93954555384[m]4.94[m]\begin{aligned} \Delta y &= 18\sin40\degree t-\frac{1}{2}gt^2\\ &= 18\sin40\degree \cdot 1.80-\frac{1}{2}g(1.80)^2\\ &=\frac{162}{5} \sin40\degree-\frac{81}{50}g\\ &\approx4.93954555384\ut{m}\\ &\approx4.94\ut{m}\\ \end{aligned}
(e) Δxt=5.00=?\Delta x_{t=5.00}=? Δx=18cos40°t=18cos40°5.00=90cos40°[m]68.9439998807[m]68.9[m]\begin{aligned} \Delta x &= 18\cos40\degree t\\ &= 18\cos40\degree \cdot 5.00\\ &=90 \cos40\degree\ut{m}\\ &\approx 68.9439998807\ut{m}\\ &\approx 68.9\ut{m}\\ \end{aligned}
(f) Δyt=5.00=?\Delta y_{t=5.00}=? Δy=18sin40°t12gt2=18sin40°5.0012g(5.00)2=90sin40°252g64.7322401282[m]64.7[m]\begin{aligned} \Delta y &= 18\sin40\degree t-\frac{1}{2}gt^2\\ &= 18\sin40\degree \cdot 5.00-\frac{1}{2}g(5.00)^2\\ &=90 \sin40\degree-\frac{25}{2}g\\ &\approx -64.7322401282\ut{m}\\ &\approx -64.7\ut{m}\\ \end{aligned}