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 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>:

$$y(t) = \frac{1}{\sqrt{2}}vt - \frac{gt^2}{2}$$

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

$$\frac{1}{\sqrt{2}}vt = L$$
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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