# Calculating Trajectories

1. Apr 11, 2006

### george01

Say in World War II a u-boat captain is trying to sink a tanker. He knows the distance to the tanker, its speed and heading (which are assumed to be constant) and the speed of the u-boats torpedos. How would he calculate what angle to fire the torpedo to hit the tanker?? (Assuming the u-boat firing calculator thing was broken and everyone else on board was busy)

2. Apr 11, 2006

### HallsofIvy

Staff Emeritus
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,
$$vt= \frac{vd\sqrt{1+ m^2}}{u}= md[/itex] That gives [tex]\sqrt{1+ m^2}= \frac{mu}{v}$$
Squaring both sides,
$$1+ m^2= \frac{m^2u^2}{v^}$$
so
$$1= \left(\frac{u^2}{v^2}-1\right)m^2$$
$$= \frac{u^2- v^2}{v^2}m^2$$
$$m^2= \frac{v^2}{u^2- v^2}$$
$$m= \frac{v}{\sqrt{u^2- v^2}}$$
so in terms of the angle:
$$\theta= arctan(\frac{v}{\sqrt{u^2- v^2}}$$
You can see why the torpedo had better be faster than the tanker!!