Updated spacing between @ latex expressions

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			<span><img src="../../img/book.png"></span><span>Savchenko Solutions</span>
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			Solutions of Savchenko Problems in Physics <br>
			<i><b>knowledge must be free</b></i>
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			<h3 id="back-link"><a href="../../#1.3">$\leftarrow$Back</a></h3>

			<h3> Statement </h3>
			<p>
			$1.3.19^*.$ What is the minimum speed required for a stone thrown by a boy to fly between height $H$ and length $L$, if the throw is made from height $h$ and the boy can choose any place to throw?
			</p>

			<h3>Solution</h3>
			<p>
			<p>The optimal trajectory, corresponding to the minimum possible throwing speed, should almost touch the edges of the house. The minimum speed of the stone at the throwing point corresponds to the minimum kinetic energy of the stone at the point of contact with the edge of the house.
			</p>
		@@ -60,7 +60,7 @@
			<p>
			Thus, the problem is reduced to determining the minimum speed of the stone at the upper corner point of the house, sufficient to cover a distance equal to the length of the house $L$ - the stone at this point should have a speed directed at an angle of $45^{ \circ}$ to the horizon. The speed of the stone at this point is determined by the known relationship:</p>
			<p class="exp">
		−	$$ L = \frac{v^{2} \sin 2\alpha }{g} $$
		+	$$ L = \frac{v^{2} \sin 2\alpha}{g} $$
			</p>
			<p>Taking into account the chosen angle, we get </p>
			<p class="exp">
			$$ v^{2} = gL $$
			</p>
			<p>The velocity at the throwing point $u$ is found from the law of conservation of energy:</p>
			<p class="exp">
			$$ \frac{mu^{2} }{2} + mgh = \frac{mv^{2} }{2} + mgH $$
			</p>
			$$u=\sqrt{g(2(H-h)+L)}$$
			</p>

			<h4>Answer</h4>
			<p>
			$$v=\sqrt{g[2(H-h)+L]}$$
			</p>


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	Solutions of Savchenko Problems in Physics <br>		Solutions of Savchenko Problems in Physics <br>
	<i><b>knowledge must be free</b></i>		<i><b>knowledge must be free</b></i>
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	<h3 id="back-link"><a href="../../#1.3">$\leftarrow$Back</a></h3>		<h3 id="back-link"><a href="../../#1.3">$\leftarrow$Back</a></h3>

	<h3> Statement </h3>		<h3> Statement </h3>
	<p>		<p>
	$1.3.19^*.$ What is the minimum speed required for a stone thrown by a boy to fly between height $H$ and length $L$, if the throw is made from height $h$ and the boy can choose any place to throw?		$1.3.19^*.$ What is the minimum speed required for a stone thrown by a boy to fly between height $H$ and length $L$, if the throw is made from height $h$ and the boy can choose any place to throw?
	</p>		</p>

	<h3>Solution</h3>		<h3>Solution</h3>
	<p>		<p>
	<p>The optimal trajectory, corresponding to the minimum possible throwing speed, should almost touch the edges of the house. The minimum speed of the stone at the throwing point corresponds to the minimum kinetic energy of the stone at the point of contact with the edge of the house.		<p>The optimal trajectory, corresponding to the minimum possible throwing speed, should almost touch the edges of the house. The minimum speed of the stone at the throwing point corresponds to the minimum kinetic energy of the stone at the point of contact with the edge of the house.
	</p>		</p>
	@@ -60,7 +60,7 @@
	<p>		<p>
	Thus, the problem is reduced to determining the minimum speed of the stone at the upper corner point of the house, sufficient to cover a distance equal to the length of the house $L$ - the stone at this point should have a speed directed at an angle of $45^{ \circ}$ to the horizon. The speed of the stone at this point is determined by the known relationship:</p>		Thus, the problem is reduced to determining the minimum speed of the stone at the upper corner point of the house, sufficient to cover a distance equal to the length of the house $L$ - the stone at this point should have a speed directed at an angle of $45^{ \circ}$ to the horizon. The speed of the stone at this point is determined by the known relationship:</p>
	<p class="exp">		<p class="exp">
	$$ L = \frac{v^{2} \sin 2\alpha }{g} $$		$$ L = \frac{v^{2} \sin 2\alpha}{g} $$
	</p>		</p>
	<p>Taking into account the chosen angle, we get </p>		<p>Taking into account the chosen angle, we get </p>
	<p class="exp">		<p class="exp">
	$$ v^{2} = gL $$		$$ v^{2} = gL $$
	</p>		</p>
	<p>The velocity at the throwing point $u$ is found from the law of conservation of energy:</p>		<p>The velocity at the throwing point $u$ is found from the law of conservation of energy:</p>
	<p class="exp">		<p class="exp">
	$$ \frac{mu^{2} }{2} + mgh = \frac{mv^{2} }{2} + mgH $$		$$ \frac{mu^{2} }{2} + mgh = \frac{mv^{2} }{2} + mgH $$
	</p>		</p>
	$$u=\sqrt{g(2(H-h)+L)}$$		$$u=\sqrt{g(2(H-h)+L)}$$
	</p>		</p>

	<h4>Answer</h4>		<h4>Answer</h4>
	<p>		<p>
	$$v=\sqrt{g[2(H-h)+L]}$$		$$v=\sqrt{g[2(H-h)+L]}$$
	</p>		</p>


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