Sounds pretty straight forward to me. Let's assume you set up a coordinate system so that the u-boat is at (0,0) and, at t=0, the tanker is on the x-axis moving parallel to the positive y-axis. If its speed is v, and distance is d, then its position at time t is (d, vt). Since the torpedo's path will be a straight line passing through (0,0), it can be written y= mx where m is the slope (tangent of the angle). But we need that in terms of t: Okay, if x= at, then y= mat and the distance along the line from the u-boat (0,0) to the torpedo is (by Pythagorean theorem) \sqrt{a^2t^2+ m^2a^2t^2}= at\sqrt{1+ m^2}. If the torpedo's speed is u, then we have ut= at\sqrt{1+ m^2} so
a= \frac{u}{\sqrt{1+ m^2}}.
The torpedo will cover the horizontal (x-direction) distance d when
at= \frac{ut}{\sqrt{1+ m^2}}= d
In otherwords, when
t= \frac{d\sqrt{1+ m^2}}{u}
The y- component at that time, y= mx= md must be equal to the distance the tanker has traveled,
vt= \frac{vd\sqrt{1+ m^2}}{u}= md[/itex]<br />
That gives<br />
\sqrt{1+ m^2}= \frac{mu}{v}<br />
Squaring both sides,<br />
1+ m^2= \frac{m^2u^2}{v^} <br />
so <br />
1= \left(\frac{u^2}{v^2}-1\right)m^2<br />
= \frac{u^2- v^2}{v^2}m^2<br />
m^2= \frac{v^2}{u^2- v^2}<br />
m= \frac{v}{\sqrt{u^2- v^2}}<br />
so in terms of the angle:<br />
\theta= arctan(\frac{v}{\sqrt{u^2- v^2}}<br />
You can see why the torpedo had better be faster than the tanker!
