Updated spacing between @ latex expressions

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revision #10440 parent #9988 GitHub c702ebe ← older newer →
en/1.3.8.md +12 −12
@@ -6,14 +6,14 @@
<meta name="viewport" content="width=device-width, initial-scale=1.0">
<meta http-equiv="content-language" content="en">
<meta name="keywords" content="Savchenko Problems in Physics, Savchenko solutions, physics problems, physics olympiad preparation, IPhO, Jaan Kalda">
− <meta name="description" content="A mortar is fired at objects located on the mountainside. At what distance from the mortar will the mines fall if their initial velocity is v, the angle of inclination of the mountain is \alpha, and the angle of fire relative to the horizon is \beta?">
+ <meta name="description" content="A mortar is fired at objects located on the mountainside. At what distance from the mortar will the mines fall if their initial velocity is v, the angle of inclination of the mountain is \alpha , and the angle of fire relative to the horizon is \beta ?">
<meta name="author" content="Aliaksandr Melnichenka">
<meta name="date" content="2023-10" scheme="YYYY-MM">
− <meta property="og:title" content="A mortar is fired at objects located on the mountainside. At what distance from the mortar will the mines fall if their initial velocity is v, the angle of inclination of the mountain is \alpha, and the angle of fire relative to the horizon is \beta?">
+ <meta property="og:title" content="A mortar is fired at objects located on the mountainside. At what distance from the mortar will the mines fall if their initial velocity is v, the angle of inclination of the mountain is \alpha , and the angle of fire relative to the horizon is \beta ?">
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− <meta property="og:description" content="A mortar is fired at objects located on the mountainside. At what distance from the mortar will the mines fall if their initial velocity is v, the angle of inclination of the mountain is \alpha, and the angle of fire relative to the horizon is \beta?">
+ <meta property="og:description" content="A mortar is fired at objects located on the mountainside. At what distance from the mortar will the mines fall if their initial velocity is v, the angle of inclination of the mountain is \alpha , and the angle of fire relative to the horizon is \beta ?">
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− <title>A mortar is fired at objects located on the mountainside. At what distance from the mortar will the mines fall if their initial velocity is v, the angle of inclination of the mountain is \alpha, and the angle of fire relative to the horizon is \beta?</title>
+ <title>A mortar is fired at objects located on the mountainside. At what distance from the mortar will the mines fall if their initial velocity is v, the angle of inclination of the mountain is \alpha , and the angle of fire relative to the horizon is \beta ?</title>
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@@ -70,28 +70,28 @@
$$ {x}={v}_{0}{t}\cos{\beta} $$
$$ {y}={v}_{0}{t}\sin{\beta}-\frac{{g}{t}^{2}}{2} $$
−Substitute into the equations the coordinates of the target $x = L; \;y = L \tan \alpha$
+Substitute into the equations the coordinates of the target $x = L; \;y = L \tan\alpha$
$$ {L=v_{0}t\cos\beta} $$
−$$ L\tan\alpha =v_{0}t\sin\beta-\frac{gt^{2}}{2} $$
+$$ L\tan\alpha =v_{0}t\sin\beta -\frac{gt^{2}}{2} $$
Let us express time from the first equation of the last system of equations and substitute its value into the second equation
$$ {t=\frac{L}{v_{0}\cos\beta}} $$
−$$ {L\tan\alpha =v_{0}\frac{L}{v_{0}\cos\beta}\sin\beta-\frac{g}{2}\frac{L^{2}}{v_{0}^{2}\cos^{2}\beta} } $$
+$$ {L\tan\alpha =v_{0}\frac{L}{v_{0}\cos\beta}\sin\beta -\frac{g}{2}\frac{L^{2}}{v_{0}^{2}\cos^{2}\beta} } $$
Where
−$$ {v}_{0}=\sqrt{\frac{{gL}\cos\alpha}{2\cos\beta\sin(\beta-\alpha)}} $$
+$$ {v}_{0}=\sqrt{\frac{{gL}\cos\alpha}{2\cos\beta\sin(\beta -\alpha )}} $$
We express $L$,
−$$ L = \frac{ 2\cos\beta\sin(\beta-\alpha)\cdot v^2_0}{g\cos\alpha} $$
+$$ L = \frac{ 2\cos\beta\sin(\beta -\alpha )\cdot v^2_0}{g\cos\alpha} $$
And we find the flight range along the wall:
−$$ l = \frac{L}{\cos \alpha} $$
+$$ l = \frac{L}{\cos\alpha} $$
−$$ \fbox{$l = \frac{ 2v^2_0}{g} \frac{ \cos\beta\sin(\beta-\alpha)}{\cos^2 \alpha}$} $$
+$$ \fbox{$l = \frac{ 2v^2_0}{g} \frac{ \cos\beta\sin(\beta -\alpha )}{\cos^2 \alpha}$} $$
</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>
unchanged lines 12