Kinematics - How fast must torpedo go

[Omid]

Here is a problem I need help with it :

A ship is steaming 45 S of W at 30 km/h. At that moment 50 km due south of it, an enemy submarine heading 30 W of N, spots it. At what speed must a torpedo from the sub travel if it's to strike the ship ? Assume that the sub fires straight ahead from its forward tubes, and overlook the fact that this problem is a bit unrealistic.

I drew a diagram and labeled it but unfortunately couldn't go any farther
Thanks

 

[Hypercase]

hey of the top of my head I am guessing something like 42.3... something kms/hr.
tell me if its right. Ill wait for u too clarify,what u mean by at what speed should the torpedo be fired? Arent those things self-propelled. I'm guessing the water ressistance will slow it down considerably.

 

[Omid]

What I wrote, is exactly what is written on my textbook except this hint :
What can be said about both westerly displacements.
And unfortunately there is no answer section for the book I am reading. So please let me know a bit about what you have done to achive the answer.
Thanks

 

[Doc Al]

What I wrote, is exactly what is written on my textbook except this hint :
What can be said about both westerly displacements.

That hint is the key. In order for the torpedo to intercept the ship, they must be at the same place at the same time. What does that tell you about the westerly component of their velocities?

 

[Omid]

But we know nothing about the submarine's velocity magnitude ! Is it still or moving ?

 

[HallsofIvy]

The submarines velocity is irrelevant. As soon as the torpedo leaves the submarine it has its own velocity: speed "v" at angle 30 degrees W of N.

Doc Al's point is that, since the torpedo's original position is exactly south of the ship (no distance east or west) the torpedo's westerly component, v sin(30), must be the same as the westerly component of the ship, (30)cos(45). Set those equal and solve for v.

 

[Omid]

Thank you very much.