10판/2. 직선운동

2-64 할리데이 10판 솔루션 일반물리학

짱세디럭스 2019. 8. 3. 16:59

$$\begin{cases} x_1=0\\ t_1=3.0\ut{s}\\ v_0=-10\ut{m/s}\\ a=-g = -9.80665\ut{m/s^2}\\ \end{cases} $$ $$ x_0=?$$ $$ \begin{aligned} \Delta x &= v_0t+\frac{1}{2}at^2,\\ x_1-x_0&= (-10)(3.0)+\frac{1}{2}(-g)(3.0)^2\\ x_0&= 30+\frac{9}{2} g\\ &= 30+\frac{9}{2}(9.80665)\\ &=\frac{2965197}{40000}\ut{m}\\ &=74.1299\ut{m}\\ &\approx 74\ut{m}\\ \end{aligned} $$