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I am obliged to count very simple problem (at least it seemed that it is simple, I hope it isn't to simple for this site). So i got observer who is standing $H$ below the object. The object is fired horizontaly with the speed $V_{0}$. Assuming that the observer has the reaction time that equals $T$, on what angle he should fire his bullet to destroy the object. We also know that bullet is fired with the speed $V_{1}$.

So I drew picture (I changed the orientation because the computation seems simplier): enter image description here

And I write equations for bullet:

$x(t)=V_{1}\cos(\alpha)\cdot t$

$y(t)=V_1\sin(\alpha)\cdot t + H - \frac{gt^2}{2}$

And for object:

$x_{1}(t)=V_{0}(t+T)$

$y_{2}(t)=\frac{g(t+T)^2}{2}$

Now I do $x(t)=x_{1}(t)$ :

and i get: $\cos (\alpha)=\frac{V_0\cdot(t+T)}{V_{1}t}$

From the second equation $y_{1}(x)=y(x)$ I count $t$ (the time they will meet) and move it to the equation $\cos (\alpha)=\frac{V_0\cdot(t+T)}{V_{1}t}$. I got very complicated and hard equations that I am not able to solve, so I am asking is it simplier way to this kind of task? Or did I miss something that makes me equations very complicated?

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  • $\begingroup$ If the computation seems simplier and is still "very complicated and hard" then I'm afraid what would it be without the change :) $\endgroup$ – Wojciech Mar 24 '14 at 19:58
  • $\begingroup$ @Wojciech Before, the computetions were very very hard, now they are very complicated :) I mean it is almost impossible the find the value for me. I mean I am not bad at maths but still it is impossible for me to solve :( I tried multiple ways, to solve this. It does not let me rest, there must be easier way to do this because Physics at my studies is not very hard, it usually comes to solving simple problems. Don't you have any suggestions? $\endgroup$ – Marcin Majewski Mar 24 '14 at 20:03
  • $\begingroup$ Maybe I have bad approach? I should start differently? $\endgroup$ – Marcin Majewski Mar 24 '14 at 20:04
  • $\begingroup$ Minor correction: you should have $y_{2}(t)=-\frac{g(t+T)^2}{2}$, not $y_{2}(t)=\frac{g(t+T)^2}{2}$. $\endgroup$ – DumpsterDoofus Mar 24 '14 at 21:20
  • $\begingroup$ @DumpsterDoofus Why? I changed the orientation so the object should fall in the same direction as gravity works. $\endgroup$ – Marcin Majewski Mar 24 '14 at 21:23

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