- #1

ozzy

- 12

- 0

Well, anyway. As one may remember, I asked about this question:

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 of the flight of the missile.

The question is - what is the time of the total flight of the missile, till it gets the aircraft?

The solution should be done without any integrals.

The solution was made by balakrishnan_v, and did I mentioned that he is THE KING?

But since then I was thinking about the solution and here what I discovered:

the relative velocity between the missile and the plane in Cartesian axis is:

Vx(t) = V*cos{f(t)} - U

Vy(t) = V*sin{f(t)} - 0

and initial position,that should preserve is R = (0 , H)

therefore:

[tex]\int_{0}^{T} (Vcos{f(t)} - U)dt = 0[/tex]

and [tex]\int_{0}^{T} (Vsin{f(t)})dt = H[/tex]

therefore [tex]\int_{0}^{T} (sin{f(t)})dt = H/V[/tex] - which is turns to be right, as checked it in MATLAB.

Though, in the axis system of the missile we get:

the relative velocity between the missile and the plane in missile axis is:

Vs(t) = V - U*cos{f(t)}

Vp(t) = 0 - (-U*sin{f(t)})

and initial position,that should preserve is R = (H , 0)

therefore:

[tex]\int_{0}^{T} (V - Ucos{f(t)})dt = H[/tex]

and [tex]\int_{0}^{T} (Usin{f(t)})dt = 0[/tex]

therefore [tex]\int_{0}^{T} (sin{f(t)})dt = 0[/tex] - Where Am I Wrong? Please help!