# Chases and Escapes: The Mathematics of Pursuit and Evasion

By Paul J. Nahin | Go to book overview

Appendix C
Solution to the Challenge
Problem of Section 1.5

To set up our analysis for this question, consider figure C1, which shows the geometry of the problem for the case 0 < θ < 45° (which means, of course, 2θ < 90°). I treat the case of θ > 45° separately. If we call the altitude of the attacking missile h and its ground distance from the defensive site d (at the instant of the anti-missile’s launch), then

If we let T denote the time interval from launch until the supposed interception, then the attacking missile flys distance VT toward the defensive site and so, at the instant of supposed interception, the attacking missile is at a ground distance of dVT. Thus,

Finally, since the defensive anti-missile missile is launched at angle 2θ, its vertical speed component is V sin(2θ); since the defensive missile must be at altitude h when interception occurs we must also have

h = VT sin(2θ)

-198-

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