Updated spacing between @ latex expressions

astrosander правка от 6 окт. 2024 г., 00:48

правка #10480 предыдущая #9578 GitHub c702ebe ← раньше позже →

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			<h3 id="back-link"><a href="../../#1.3">$\leftarrow$Back</a></h3>

			<h3> Statement </h3>
			<p>
			$1.3.5.$ A stone is thrown at a velocity $v$ at an angle $\varphi$ to the horizon. After what time will the velocity be at the angle $\alpha$ with the horizon?
			</p>

		@@ -56,7 +56,7 @@
			<h3>Solution</h3>
			<p>
			The horizontal component of velocity remains unchanged:
		−	$$ v_{x}(\varphi) = v_{x}(\alpha) = v \cdot \cos{\varphi} $$
		+	$$ v_{x}(\varphi) = v_{x}(\alpha ) = v \cdot \cos{\varphi} $$
			And the horizontal component decreases, depending on time, according to the law:
			$$ v_{y}(t) = vt \sin{\varphi} - gt $$
			From where the angle that the velocity makes with the horizon is determined as:
			$$ \tan{\alpha} = \frac{v_{y}(t)}{v_x} $$
			Or,
			$$ \tan{\alpha} \cdot v \cdot \cos{\varphi} = v \cdot \sin{\varphi} - gt $$
			From where we obtain the required moment of time:
			$$ \fbox{$t= \frac{v}{g}(\sin{\varphi} - \cos{\varphi}\tan{\alpha})$} $$

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			<h4>Answer</h4>
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			$$t= \frac{v}{g}(\sin{\varphi} - \cos{\varphi}\tan{\alpha})$$
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	<h3> Statement </h3>		<h3> Statement </h3>
	<p>		<p>
	$1.3.5.$ A stone is thrown at a velocity $v$ at an angle $\varphi$ to the horizon. After what time will the velocity be at the angle $\alpha$ with the horizon?		$1.3.5.$ A stone is thrown at a velocity $v$ at an angle $\varphi$ to the horizon. After what time will the velocity be at the angle $\alpha$ with the horizon?
	</p>		</p>

	@@ -56,7 +56,7 @@
	<h3>Solution</h3>		<h3>Solution</h3>
	<p>		<p>
	The horizontal component of velocity remains unchanged:		The horizontal component of velocity remains unchanged:
	$$ v_{x}(\varphi) = v_{x}(\alpha) = v \cdot \cos{\varphi} $$		$$ v_{x}(\varphi) = v_{x}(\alpha ) = v \cdot \cos{\varphi} $$
	And the horizontal component decreases, depending on time, according to the law:		And the horizontal component decreases, depending on time, according to the law:
	$$ v_{y}(t) = vt \sin{\varphi} - gt $$		$$ v_{y}(t) = vt \sin{\varphi} - gt $$
	From where the angle that the velocity makes with the horizon is determined as:		From where the angle that the velocity makes with the horizon is determined as:
	$$ \tan{\alpha} = \frac{v_{y}(t)}{v_x} $$		$$ \tan{\alpha} = \frac{v_{y}(t)}{v_x} $$
	Or,		Or,
	$$ \tan{\alpha} \cdot v \cdot \cos{\varphi} = v \cdot \sin{\varphi} - gt $$		$$ \tan{\alpha} \cdot v \cdot \cos{\varphi} = v \cdot \sin{\varphi} - gt $$
	From where we obtain the required moment of time:		From where we obtain the required moment of time:
	$$ \fbox{$t= \frac{v}{g}(\sin{\varphi} - \cos{\varphi}\tan{\alpha})$} $$		$$ \fbox{$t= \frac{v}{g}(\sin{\varphi} - \cos{\varphi}\tan{\alpha})$} $$

	</p>		</p>

	<h4>Answer</h4>		<h4>Answer</h4>
	<p>		<p>
	$$t= \frac{v}{g}(\sin{\varphi} - \cos{\varphi}\tan{\alpha})$$		$$t= \frac{v}{g}(\sin{\varphi} - \cos{\varphi}\tan{\alpha})$$
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