Updated greek laters @ latex compiling

astrosander edited
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revision #9988 parent #9594 GitHub dbb31ba ← older newer →
en/1.3.8.md +4 −4
@@ -73,25 +73,25 @@
Substitute into the equations the coordinates of the target $x = L; \;y = L \tan \alpha$
$$ {L=v_{0}t\cos\beta} $$
−$$ Ltg\alpha=v_{0}t\sin\beta-\frac{gt^{2}}{2} $$
+$$ L\tan\alpha =v_{0}t\sin\beta-\frac{gt^{2}}{2} $$
Let us express time from the first equation of the last system of equations and substitute its value into the second equation
$$ {t=\frac{L}{v_{0}\cos\beta}} $$
−$$ {Ltg\alpha=v_{0}\frac{L}{v_{0}\cos\beta}sin\beta-\frac{g}{2}\frac{L^{2}}{v_{0}^{2}\cos^{2}\beta} } $$
+$$ {L\tan\alpha =v_{0}\frac{L}{v_{0}\cos\beta}\sin\beta-\frac{g}{2}\frac{L^{2}}{v_{0}^{2}\cos^{2}\beta} } $$
Where
$$ {v}_{0}=\sqrt{\frac{{gL}\cos\alpha}{2\cos\beta\sin(\beta-\alpha)}} $$
We express $L$,
$$ L = \frac{ 2\cos\beta\sin(\beta-\alpha)\cdot v^2_0}{g\cos\alpha} $$
And we find the flight range along the wall:
$$ l = \frac{L}{\cos \alpha} $$
−$$ \fbox{$l = \frac{ 2v^2_0}{g} \frac{ \cos\beta\sin(\beta-\alpha)}{\cos^2\alpha}$} $$
+$$ \fbox{$l = \frac{ 2v^2_0}{g} \frac{ \cos\beta\sin(\beta-\alpha)}{\cos^2 \alpha}$} $$
</p>
<h4>Answer</h4>
<p>
− $$L=\frac{2v^2}g\frac{\cos^2\beta}{\cos\alpha}(\text{tg}\beta-\text{tg}\alpha)$$
+ $$L=\frac{2v^2}g\frac{\cos^2 \beta}{\cos\alpha}(\text{tg}\beta-\text{tg}\alpha)$$
</p>
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