Updated greek laters @ latex compiling

astrosander правка от
Посмотреть решение История
правка #9969 предыдущая #9522 GitHub dbb31ba ← раньше позже →
en/1.3.27.md +5 −5
@@ -66,21 +66,21 @@
</p>
<p>The maximum height of the stone lift along the vertical axis should be equal to $y_{max} = 2R$, therefore</p>
<p class="exp">
−$$ \frac{v_0^2 \sin^2 \alpha}{2g} = 2R $$
+$$ \frac{v_0^2 \sin^2 \alpha}{2g} = 2R $$
</p>
<p>Determine the value of the initial throw speed</p>
<p class="exp">
−$$ v_0 = \sqrt{\frac{4gR}{\sin^2 \alpha}} $$
+$$ v_0 = \sqrt{\frac{4gR}{\sin^2 \alpha}} $$
</p>
<p>The angle $\alpha$ at which the stone should be thrown is determined from the initial conditions</p>
<p class="exp">
−$$ v_0t_1 \cos \alpha = \frac{v_0^2 \sin \alpha \cos \alpha}{g}$$
+$$ v_0t_1 \cos \alpha = \frac{v_0^2 \sin \alpha \cos \alpha}{g}$$
$$ 2R = \frac{v_0^2 \sin \alpha \cos \alpha}{g}$$
−$$\tan \alpha = 2; \quad \alpha = \text{arctg} \;2 \approx 63^\circ$$
+$$\tan \alpha = 2; \quad \alpha = \text{\arctan } \;2 \approx 63^\circ$$
</p>
<p>Substituting into the formula for $v_0$</p>
<p class="exp">
−$$ \fbox{$v_0 = \sqrt{\frac{4gR}{\sin^2 63^\circ}} = \sqrt{5Rg}$} $$
+$$ \fbox{$v_0 = \sqrt{\frac{4gR}{\sin^2 63^\circ}} = \sqrt{5Rg}$} $$
</p>
</p>
ещё строк без изменений 18