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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>
			$1.3.9.$ At what speed should a projectile fly out of a cannon at the moment of rocket launch in order to hit a rocket starting vertically with acceleration $a$? The distance from the gun to the rocket launch site is $L$, the gun fires at an angle of $45^\circ$ to the horizon.
			</p>

			<h3>Solution</h3>
			<p>
			Let's use the formula for the $x$ and $y$ coordinates obtained in <a href="../1.3.6" target="_blank">1.3.6</a>:
			$$ y(t) = \frac{1}{\sqrt{2}}vt - \frac{gt^2}{2} $$

			$$ x(t) = \frac{1}{\sqrt{2}}vt $$
		@@ -62,7 +62,7 @@
			In this case, the equation describing the rocket’s motion is:
			$$ y(t) = \frac{at^2}{2} $$
			We write down the meeting conditions:
		−	$$ \frac{1}{\sqrt{2}}vt - \frac{gt^2}{2} = \frac{at^2}{2}\;(1) $$
		+	$$ \frac{1}{\sqrt{2}}vt - \frac{gt^2}{2} = \frac{at^2}{2}\quad(1) $$

			$$ \frac{1}{\sqrt{2}}vt = L $$
			Where, moment of meeting:
			$$ t = \frac{\sqrt{2}L}{v} $$

			$$ L=\frac{v\sqrt{2}}{g} $$
			Substitute into $(1)$:
			$$ \frac{v\sqrt{2}}{g} = \frac{(a+g)v^2}{g} $$
			We obtain the required speed:
			$$ \fbox{$ v=\sqrt{L(a+g)} $} $$

			</p>

			<h4>Answer</h4>
			<p>
			$$v=\sqrt{L(a+g)}$$
			</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>
	$1.3.9.$ At what speed should a projectile fly out of a cannon at the moment of rocket launch in order to hit a rocket starting vertically with acceleration $a$? The distance from the gun to the rocket launch site is $L$, the gun fires at an angle of $45^\circ$ to the horizon.		$1.3.9.$ At what speed should a projectile fly out of a cannon at the moment of rocket launch in order to hit a rocket starting vertically with acceleration $a$? The distance from the gun to the rocket launch site is $L$, the gun fires at an angle of $45^\circ$ to the horizon.
	</p>		</p>

	<h3>Solution</h3>		<h3>Solution</h3>
	<p>		<p>
	Let's use the formula for the $x$ and $y$ coordinates obtained in <a href="../1.3.6" target="_blank">1.3.6</a>:		Let's use the formula for the $x$ and $y$ coordinates obtained in <a href="../1.3.6" target="_blank">1.3.6</a>:
	$$ y(t) = \frac{1}{\sqrt{2}}vt - \frac{gt^2}{2} $$		$$ y(t) = \frac{1}{\sqrt{2}}vt - \frac{gt^2}{2} $$

	$$ x(t) = \frac{1}{\sqrt{2}}vt $$		$$ x(t) = \frac{1}{\sqrt{2}}vt $$
	@@ -62,7 +62,7 @@
	In this case, the equation describing the rocket’s motion is:		In this case, the equation describing the rocket’s motion is:
	$$ y(t) = \frac{at^2}{2} $$		$$ y(t) = \frac{at^2}{2} $$
	We write down the meeting conditions:		We write down the meeting conditions:
	$$ \frac{1}{\sqrt{2}}vt - \frac{gt^2}{2} = \frac{at^2}{2}\;(1) $$		$$ \frac{1}{\sqrt{2}}vt - \frac{gt^2}{2} = \frac{at^2}{2}\quad(1) $$

	$$ \frac{1}{\sqrt{2}}vt = L $$		$$ \frac{1}{\sqrt{2}}vt = L $$
	Where, moment of meeting:		Where, moment of meeting:
	$$ t = \frac{\sqrt{2}L}{v} $$		$$ t = \frac{\sqrt{2}L}{v} $$

	$$ L=\frac{v\sqrt{2}}{g} $$		$$ L=\frac{v\sqrt{2}}{g} $$
	Substitute into $(1)$:		Substitute into $(1)$:
	$$ \frac{v\sqrt{2}}{g} = \frac{(a+g)v^2}{g} $$		$$ \frac{v\sqrt{2}}{g} = \frac{(a+g)v^2}{g} $$
	We obtain the required speed:		We obtain the required speed:
	$$ \fbox{$ v=\sqrt{L(a+g)} $} $$		$$ \fbox{$ v=\sqrt{L(a+g)} $} $$

	</p>		</p>

	<h4>Answer</h4>		<h4>Answer</h4>
	<p>		<p>
	$$v=\sqrt{L(a+g)}$$		$$v=\sqrt{L(a+g)}$$
	</p>		</p>


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