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.
Solution
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: $$\frac{1}{\sqrt{2}}vt - \frac{gt^2}{2} = \frac{at^2}{2}\quad(1)$$$$\frac{1}{\sqrt{2}}vt = L$$ 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: $$\fbox{$ v=\sqrt{L(a+g)} $}$$