$1.3.9.$ 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.
Let's use the formula for the $x$ and $y$ coordinates obtained in 1.3.6: $$ y(t) = \frac{1}{\sqrt{2}}vt - \frac{gt^2}{2} $$ $$ x(t) = \frac{1}{\sqrt{2}}vt $$ 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: $$ \fbox{$ v=\sqrt{L(a+g)} $} $$
$$v=\sqrt{L(a+g)}$$