솔루션피아

(풀이자주 : 속력이 m단위라 임의로 m/s로 간주하여 풀었습니다.) $$\begin{cases} \theta_0 &= 30\degree\\ v_0 &= 25.0\ut{m/s}\\ d &= 20.0\ut{m}\\ \end{cases}$$ $$\begin{cases} \vec a &=-g\j\\ g &\approx 9.80665\ut{m/s^2}\\ \end{cases}$$ $$\begin{aligned} \Delta t_{0\rarr1} &= \frac{\Delta x_{0\rarr1}}{v_x}\\ &=\frac{d}{v_0\cos\theta}\\ &=\frac{8}{5 \sqrt{3}} \end{aligned}$$ (a) $h_1=?$ $$\Delta r = v_{0}t+\frac{1}{2}at^2,..

$$\begin{cases} \vec r_Q &= (32,3.5)\ut{m}\\ \vec r_P &= (0,1.0)\ut{m}\\ \vec v_0 &= (18\ut{m/s},40\degree)\\ \vec a &=-g\j\\ g &\approx 9.80665\ut{m/s^2}\\ \end{cases}$$ $$\title{Over Q Point Possible?}$$ $$v_0 = 18\cos40\degree\i+18\sin40\degree\j,$$ $$\begin{aligned} t &= \frac{\Delta x}{v_x}\\ &= \frac{32}{18\cos40\degree}\\ &= \frac{16}{9} \sec40\degree\\ \end{aligned}$$ $$\begin{aligned} \..

$$\begin{cases} \vec v_{A0} &= (23.1\ut{m/s},45\degree)\\ x_{B0} &= 55\ut{m}\\ \vec a &=-g\j\\ g &\approx 9.80665\ut{m/s^2}\\ \end{cases}$$ $$\bar v_B=?$$ $$\begin{aligned} \vec v_{A0} &= (23.1\ut{m/s},45\degree)\\ &= \frac{231}{10 \sqrt{2}}\i+\frac{231}{10 \sqrt{2}}\j\\ \end{aligned}$$ $$\begin{aligned} \Delta y_A&=v_{Ay0}t+\frac{1}{2}a_yt^2,\\ 0 &= v_{Ay0}t+\frac{1}{2}(-g)t^2\\ &= v_{Ay0}-\fra..

$$\begin{cases} v_0 &= 6.00v_{H}\\ \vec a &= -g\j\\ \end{cases}$$ $$\theta_0=?$$ $$\begin{aligned} \vec v_H &= \vec v_x,\\ \end{aligned}$$ $$\begin{aligned} \theta_0&=\cos^{-1}\frac{v_x}{v_0}\\ &=\cos^{-1}\frac{v_H}{6v_H}\\ &=\cos^{-1}\frac{1}{6}\\ &\approx 1.40334824758\ut{rad}\\ &\approx 1.40\ut{rad}\\ \end{aligned}$$

$$\begin{cases} \vec v_0 &= (42.0\ut{m/s},60.0\degree)\\ t &= 5.50\ut{s}\\ \vec a &= -g\j\\ g &\approx 9.80665\ut{m/s^2}\\ \end{cases}$$ $$\begin{aligned} \vec v_0 &= (42.0\ut{m/s},60.0\degree)\\ &=21\i+21\sqrt{3}\j\ut{m/s}\\ \end{aligned}$$ (a) $h=?$ $$\begin{aligned} \Delta y &= v_{y0}t+\frac{1}{2}at^2,\\ h &= \(21\sqrt{3}\)(5.50)+\frac{1}{2}(-g)(5.50)^2\\ &=\frac{231 }{2}\sqrt{3}-\frac{121 }{..

$$\begin{cases} \vec v_0 &= (290.0\ut{km/h},-30.0\degree)\\ \Delta x &= 700\ut{m}\\ \vec a &= -g\j\\ g &\approx 9.80665\ut{m/s^2}\\ \end{cases}$$ $$\begin{aligned} \vec v_0 &= (290.0\ut{km/h},-30.0\degree)\\ &=(\frac{725}{9}\ut{m/s},-30.0\degree)\\ &=\frac{725}{6 \sqrt{3}}\i-\frac{725}{18}\j\\ \end{aligned}$$ (a) $t=?$ $$\begin{aligned} \Delta x &= v_x t,\\ t&=\frac{\Delta x}{v_x}\\ &=\frac{700}..

$$\begin{cases} \vec v_0 &= (18.0\ut{m/s},40.0\degree)\\ \vec a &= -g\j\\ g &\approx 9.80665\ut{m/s^2}\\ \end{cases}$$ $$\begin{aligned} \vec v_0 &= (18.0\ut{m/s},40.0\degree)\\ &=18\cos40\degree\i+ 18\sin40\degree\j\\ \end{aligned}$$ $$\begin{aligned} \Delta y &= v_{y0}t+\frac{1}{2}a_yt^2,\\ &= 18\sin40\degree t+\frac{1}{2}(-g)t^2\\ &= 18\sin40\degree t-\frac{1}{2}gt^2\\ \end{aligned}$$ $$ \vec..

$$\begin{cases} \vec a &= -g\j\\ g &\approx 9.80665\ut{m/s^2}\\ \end{cases}$$ $$\begin{cases} \Delta x &= 77.0\ut{m}\\ \Delta y &= 0\\ \theta &= 12.0\degree\\ \end{cases}$$ $$\begin{aligned} \Delta y &= v_{y0}t+\frac{1}{2}a_yt^2,\\ 0 &= (v_0\sin\theta) t+\frac{1}{2}(-g)t^2\\ 0 &= v_0\sin\theta -\frac{1}{2}gt\\ t &= \frac{2v_0}{g}\sin\theta\\ \end{aligned}$$ $$\begin{aligned} \Delta x &= v_xt,\\ ..

$$\begin{cases} \vec a &= -g\j\\ g &= 9.8\ut{m/s^2}\\ \end{cases}$$ $$\begin{cases} \Delta x &= 8.95\ut{m}\\ v_0 &= 9.5\ut{m/s}\\ \end{cases}$$ $$\begin{cases} v_x &= v_0\cos\theta\\ v_y &= v_0\sin\theta\\ \end{cases}$$ $$\begin{aligned} \Delta y &= v_{y0}t+\frac{1}{2}a_yt^2,\\ 0 &= (v_0\sin\theta) t+\frac{1}{2}(-g)t^2\\ 0 &= v_0\sin\theta -\frac{1}{2}gt\\ t &= \frac{2v_0}{g}\sin\theta\\ \end{al..

$$\begin{cases} \vec a &= -g\j\\ g &\approx 9.80665\ut{m/s^2}\\ \end{cases}$$ $$\begin{cases} h &= 40.4\ut{m}\\ \vec v_0 &= 285\i\ut{m/s}\\ \end{cases}$$ (a) $t=?$ $$\begin{aligned} \Delta y &= v_{y0}t+\frac{1}{2}a_yt^2,\\ -h &= (0)t+\frac{1}{2}(-g)t^2\\ h &= \frac{1}{2}gt^2\\ t &=\sqrt{\frac{2h}{g}}\\ &\approx2.8704193075\ut{s}\\ &\approx2.87\ut{s} \end{aligned}$$ (b) $\Delta x=?$ $$\begin{alig..