# 牛顿力学 Newtonian Mechanics MATH122

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$$\begin{array}{ll} v_{x}=v_{0 x}+a_{x} t & v_{y}=v_{0_{y}}+a_{y} t \ v_{x}=v_{0} \cos \theta_{0} & v_{y}=v_{0} \sin \theta_{0}+(-g) t \end{array}$$
Then from these components we havc
Magnitude of $v: \quad v=\left(v_{x}^{2}+v_{v}\right)^{1 / 2}$
Direction oi’ y: $\quad \tan \theta=\frac{v_{x}}{v_{x}}$

## MATH122COURSE NOTES ：

Therefore,
$$L=v t$$
Vertical component:
\begin{aligned} &y=v_{0 y} t+\frac{1}{2} a_{y} t^{2} \ &y=0+\frac{1}{2}(-g) t^{2} \end{aligned}
which upon substitution for $t$ gives us
$$y=\frac{-g}{2}\left(\frac{L}{v}\right)^{2}$$
Suppose that we apply this result to a beam of atoms with a speed of about $500 \mathrm{~m} / \mathrm{sec}$. In traveling a horizontal distance of $1 \mathrm{~m}$, the time of flight would be $1 / 500 \mathrm{sec}(2 \mathrm{msec})$ and we should have
$$y=-\frac{9.8}{2}\left(\frac{1}{500}\right)^{2} \approx-2 \times 10^{-5} \mathrm{~m}$$