4-21 할리데이 11판 솔루션 일반물리학

$$\begin{cases} y_0&=0.762\ut{m}\\ v_0&=33.53\ut{m/s}\\ \theta_0&=55.0\degree\\ t&=5\ut{s}\\ h&=11.28\ut{m}\\ g&=9.80665\ut{m/s^2}\\ \end{cases}$$ $$\begin{aligned} \vec v &= v_0\cos\theta_0\i+v_0\sin\theta_0\j\\ &=33.53\cos55\degree\i+33.53\sin55\degree\j\\ \end{aligned}$$ $$\ab{a}$$ $$\begin{aligned} \Delta x_5 &= v_x t\\ &=33.53\cos55\degree \cdot 5\\ &\approx 96.16008955425288\ut{m}\\ &\approx 96.2\ut{m}\\ \end{aligned}$$ $$\ab{b}$$ $$S=v_0t+\frac{1}{2}at^2,$$ $$\begin{aligned} \Delta y_5&=(33.53\sin55\degree)\cdot5+\frac{1}{2}(-g)(5)^2\\ &\approx 14.74771522504949\ut{m}\\ &\approx 14.7\ut{m}\\ \end{aligned}$$ $$\ab{c}$$ $$\begin{aligned} \Delta x_5 &= v_x t\\ &=33.53\cos55\degree \cdot 4\\ &\approx 76.9280716434023\ut{m}\\ &\approx 76.9\ut{m}\\ \end{aligned}$$ $$\ab{d}$$ $$\begin{aligned} \Delta x_4&=(33.53\sin55\degree)\cdot4+\frac{1}{2}(-g)(4)^2\\ &=31.41147218003958\ut{m}\\ \end{aligned}$$