Updated spacing between @ latex expressions

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правка #10601 предыдущая #9969 GitHub c702ebe ← раньше позже →
en/1.3.27.md +6 −6
@@ -57,12 +57,12 @@
<p>
<p>The stone must be thrown at an angle $\alpha$ to the horizon, satisfying the equations obtained in <a href="../1.3.6" target="_blank">1.3.6</a>:</p>
<p class="exp">
−$$v_x = v \cos \alpha; \quad v_y = v \sin \alpha - gt;$$
−$$x = vt \cos \alpha; \quad y = vt \sin \alpha - gt^2 / 2.$$
+$$v_x = v \cos\alpha ; \quad v_y = v \sin\alpha - gt;$$
+$$x = vt \cos\alpha ; \quad y = vt \sin\alpha - gt^2 / 2.$$
</p>
<p>The time it takes for the stone to rise to the maximum height $2R$ is found as</p>
<p class="exp">
−$$ t_1 = \frac{v_0 \sin \alpha}{g} $$
+$$ t_1 = \frac{v_0 \sin\alpha}{g} $$
</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">
@@ -74,9 +74,9 @@
</p>
<p>The angle $\alpha$ at which the stone should be thrown is determined from the initial conditions</p>
<p class="exp">
−$$ v_0t_1 \cos \alpha = \frac{v_0^2 \sin \alpha \cos \alpha}{g}$$
−$$ 2R = \frac{v_0^2 \sin \alpha \cos \alpha}{g}$$
−$$\tan \alpha = 2; \quad \alpha = \text{\arctan } \;2 \approx 63^\circ$$
+$$ v_0t_1 \cos\alpha = \frac{v_0^2 \sin\alpha\cos\alpha}{g}$$
+$$ 2R = \frac{v_0^2 \sin\alpha\cos\alpha}{g}$$
+$$\tan\alpha = 2; \quad \alpha = \text{\arctan } \;2 \approx 63^\circ$$
</p>
<p>Substituting into the formula for $v_0$</p>
<p class="exp">
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