Правка раздела «Solution»
en/1.3.9.md
+1 −1
| ### Statement | |||
| $1.3.9.$ 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. | |||
| @@ -4,8 +4,8 @@Statement | |||
| ### Solution | |||
| − | Let's use the formula for the $x$ and $y$ coordinates obtained in [1.3.6](../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: $$ \frac{1}{\sqrt{2}}vt - \frac{gt^2}{2} = \frac{at^2}{2} | ||
| + | Let's use the formula for the $x$ and $y$ coordinates obtained in [1.3.6](../1.3.6): $$ 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}\tag{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)} $} $$ | ||
| #### Answer | |||
| $$v=\sqrt{L(a+g)}$$ | |||
| ### Statement | ### Statement | ||
| $1.3.9.$ At what speed should a projectile fly out of a cannon at the moment of rocket launch in order to hit a rocket starting vertically with acceleration $a$? The distance from the gun to the rocket launch site is $L$, the gun fires at an angle of $45^\circ$ to the horizon. | $1.3.9.$ At what speed should a projectile fly out of a cannon at the moment of rocket launch in order to hit a rocket starting vertically with acceleration $a$? The distance from the gun to the rocket launch site is $L$, the gun fires at an angle of $45^\circ$ to the horizon. | ||
| @@ -4,8 +4,8 @@Statement | |||
| ### Solution | ### Solution | ||
| Let's use the formula for the $x$ and $y$ coordinates obtained in [1.3.6](../1.3.6): $$ 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} |
Let's use the formula for the $x$ and $y$ coordinates obtained in [1.3.6](../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: $$ \frac{1}{\sqrt{2}}vt - \frac{gt^2}{2} = \frac{at^2}{2}\tag{1} $$ $$ \frac{1}{\sqrt{2}}vt = L $$ 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: $$ \fbox{$ v=\sqrt{L(a+g)} $} $$ | ||
| #### Answer | #### Answer | ||
| $$v=\sqrt{L(a+g)}$$ | $$v=\sqrt{L(a+g)}$$ | ||