I have the following problem:

An airplane P, flying at a height of $h$ needs to hit the target on the ground $T$. The airplane is flying horizontaly at a constant speed $V_0$.
Find the Cauchy problem that $x()$ and $y()$ must satisfy, knowing that only gravity acts on the projectile.
At what distance $x^*$ must the projectile be launched in order to hit the target? After what amount of time $t^*$ will the projectile hit the target?
I know a little about projectile motion and I am very confused at the part where the Cauchy problem is asked? What is that asking? Is it asking to separate the horizontal and vertical movement in two function $x$ and $y$ ? I don't want you to solve my problem for me but any hint would be very muh appreciated. Thank you!


The way I interpret the problem it the following. Consider the regular x-y plane. At time $t=0$ we have a projectile at $x=0$ and $y=h$. Since it also has the same velocity of the plane it will move in the positive $x$-direction with velocity $V_0$ when $t \geq 0$. At $t=0$ we also know that the projectile is at $y=h$ and has zero velocity. However it will get some velocity in the negative $y$-direction when the time starts running, since gravity is acting on it.

Now it is you task to find the $x$ and $y$ coordinate of the projectile for $t > 0$. Thus filling in $$ \begin{aligned} y(t) &= ? \\ x(t) &= ?\end{aligned}$$ Taking into account the things we know from the situation at $t=0$. Can you do this?

  • $\begingroup$ It would be $\left\{\begin{matrix} x'(t)=V_0\cos\theta & \\ y'(t)=V_0\sin\theta -gt& \end{matrix}\right. $ wih initial conditions $x(t) = 0$ and $y(t) = h$ $\endgroup$ – Raducu Mihai May 27 at 14:45
  • $\begingroup$ I wasn’t aware of the angle, but yeah this is basically it! Can you now solve the problem? $\endgroup$ – Tim Dikland May 27 at 15:27
  • $\begingroup$ I think so. $x(t) = V_0\cos\theta t$ and $y(t) = V_0\sin\theta t + \frac{gt/2}{2}+h$ $\endgroup$ – Raducu Mihai May 27 at 22:12
  • $\begingroup$ Not quite. You didn't integrate correctly and think about the signs. $x(t) = V_0 \cos (\theta) t$ and $y(t) = -V_0 \sin (\theta)t - \frac{1}{2}gt^2 + h$ $\endgroup$ – Tim Dikland May 29 at 9:30
  • $\begingroup$ Yep, now I got the idea. Thank you very much for your help! $\endgroup$ – Raducu Mihai Jun 3 at 13:10

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