10판/4. 2차원운동과 3차원운동

4-40 할리데이 10판 솔루션 일반물리학

짱세디럭스 2021. 3. 26. 15:18
{H=2.160[m]v0=15.00[m/s]\begin{cases} H &= 2.160\ut{m}\\ v_0 &= 15.00\ut{m/s}\\ \end{cases} {a=gj^g9.80665[m/s2]\begin{cases} \vec a &=-g\j\\ g &\approx 9.80665\ut{m/s^2}\\ \end{cases} Δy=v0yt+12at2H=v0tsinθ12gt2(2.160)=(15.00)tsinθ12gt2\begin{aligned} \Delta y&=v_{0y}t+\frac{1}{2}at^2\\ -H&=v_0t\sin\theta-\frac{1}{2}gt^2\\ -(2.160)&=(15.00)t\sin\theta-\frac{1}{2}gt^2\\ \end{aligned} t=35g(12g+625sin2θ+25sinθ) \therefore t=\frac{3}{5 g} \left(\sqrt{12 g+625 \sin ^2\theta}+25 \sin \theta\right) Δx=vxt=v0cosθt=(15.00)cosθ35g(12g+625sin2θ+25sinθ)=9g(12g+625sin2θ+25sinθ)cosθ\begin{aligned} \Delta x &= v_x t\\ &=v_0\cos\theta \cdot t\\ &=(15.00)\cos\theta \cdot \frac{3}{5 g} \left(\sqrt{12 g+625 \sin ^2\theta}+25 \sin \theta\right)\\ &=\frac{9 }{g}\left(\sqrt{12 g+625 \sin ^2\theta}+25 \sin \theta\right)\cos \theta \end{aligned}
(a) θ0=45.0°,Δx=?\theta_0=45.0\degree, \Delta x=? Δx=9g(12g+625sin2θ+25sinθ)cosθ=92g(24g+625+25)24.9313977643[m]24.9[m]\begin{aligned} \Delta x &=\frac{9}{g}\left(\sqrt{12 g+625 \sin ^2\theta}+25 \sin \theta\right)\cos \theta\\ &= \frac{9 }{2 g}\left(\sqrt{24 g+625}+25\right)\\ &\approx 24.9313977643\ut{m}\\ &\approx 24.9\ut{m}\\ \end{aligned}
(b) θ0=42.0°,Δx=?\theta_0=42.0\degree, \Delta x=? Δx=9g(12g+625sin2θ+25sinθ)cosθ25.0068642996[m]25.0[m]\begin{aligned} \Delta x &=\frac{9}{g}\left(\sqrt{12 g+625 \sin ^2\theta}+25 \sin \theta\right)\cos \theta\\ &\approx 25.0068642996\ut{m}\\ &\approx 25.0\ut{m}\\ \end{aligned}