An aircraft flights at height H with a constant velocity U. When it passes right above the missile, the missile is fired with a constant velocity V (V>U), but the vector of the velocity of the missile points directly to the aircraft all the time during the flight of the missile, up till it hits the aircraft. The question is - what is the time of the total flight of the missile, right before it gets the aircraft? The solution should be done without any integrals.

• The velocity of the missile in observational frame of reference is $Vcos left( {varphi left( t right)} right)$ in horizontal direction, and $Vsin left( {varphi left( t right)} right)$ in vertical direction, where $varphi left( t right)$ is the instantaneous angle the missile is making with the horizontal axis.
  In horizontal direction both the missile and the aircraft has to travel the same distance before they meet, therefore .
  The aircraft velocity, as it seen in the missile frame of reference is in horizontal direction only and is equal to . The distance that the plane has to “travel” in missile frame of reference is the distance H (since this is the initial distance between the missile nose and the plane). Therefore $intlimits_0^T {cos left( {varphi left( t right)} right)dt} = frac{{UT}}{V}$ . Substituting the previously obtained integral, we get: $left| {frac{{{U^2}T}}{V} - VT} right| = H$. And finally, with a little algebra we obtain $T = frac{{HV}}{{{V^2} - {U^2}}}$ .
  
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