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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> |
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| <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. |
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| <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>: |
| $$ y(t) = \frac{1}{\sqrt{2}}vt - \frac{gt^2}{2} $$ | | $$ y(t) = \frac{1}{\sqrt{2}}vt - \frac{gt^2}{2} $$ |
| | | |
| $$ x(t) = \frac{1}{\sqrt{2}}vt $$ | | $$ x(t) = \frac{1}{\sqrt{2}}vt $$ |
| 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: |
| $$ \frac{1}{\sqrt{2}}vt - \frac{gt^2}{2} = \frac{at^2}{2}\;(1) $$ | | $$ \frac{1}{\sqrt{2}}vt - \frac{gt^2}{2} = \frac{at^2}{2}\;(1) $$ |
| | | |
| $$ \frac{1}{\sqrt{2}}vt = L $$ | | $$ \frac{1}{\sqrt{2}}vt = L $$ |
| 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: |
| $$ \fbox{$ v=\sqrt{L(a+g)} $} $$ | | $$ \fbox{$ v=\sqrt{L(a+g)} $} $$ |
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| </p> | | </p> |
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| <h4>Answer</h4> | | <h4>Answer</h4> |
| <p> | | <p> |
| $$v=\sqrt{L(a+g)}$$ | | $$v=\sqrt{L(a+g)}$$ |
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