Updated spacing between @ latex expressions

astrosander правка от
правка #10384 предыдущая #9947 GitHub c702ebe ← раньше позже →
@@ -79,24 +79,24 @@
</center>
<p>
Let's consider this equation of motion:
−$$ y = h = \nu \sin \alpha - \frac{gt^{2}}{2} $$
+$$ y = h = \nu\sin\alpha - \frac{gt^{2}}{2} $$
It is known that:
$$ \left\{\begin{matrix}
S_{yt}=ut \\
−S_{o}=\nu t \cos \alpha
+S_{o}=\nu t \cos\alpha
\end{matrix}\right. $$
For angle $\alpha$:
−$$ \tan \alpha = \frac{h}{S_{o} - S_{yt}} = \frac{h}{t(\nu \cos \alpha - u)} $$
+$$ \tan\alpha = \frac{h}{S_{o} - S_{yt}} = \frac{h}{t(\nu\cos\alpha - u)} $$
We substitute the value of the angle $\alpha$ into the equation for the height:
−$$ h = t (\nu \cos \alpha - u) \tan \alpha = \nu t \sin \alpha - \frac{gt^{2}}{2} $$
+$$ h = t (\nu\cos\alpha - u) \tan\alpha = \nu t \sin\alpha - \frac{gt^{2}}{2} $$
From this equation we can express time $t$:
−$$ t = \frac{2u \tan \alpha}{g} $$
+$$ t = \frac{2u \tan\alpha}{g} $$
We substitute the time value $t$ back into the equation for height:
−$$ \fbox{$h = \frac{2u \tan^{2} \alpha}{g} (\nu \cos \alpha - u)$} $$
+$$ \fbox{$h = \frac{2u \tan^{2} \alpha}{g} (\nu\cos\alpha - u)$} $$
<h4>Answer</h4>
<p>
− $$h=\frac{2utg^{2}\alpha}{g}(\nu \cos\alpha -u)$$
+ $$h=\frac{2utg^{2}\alpha}{g}(\nu\cos\alpha -u)$$
</p>
<p style="text-align: right; font-style: italic; font-size: 14;">
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