Finding Theta: Aiming a Gun Between Two Ships

[teddyayalew]

Homework Statement

Two ships are traveling parallel to each other in opposite directions with seed v1 and v2. One ship fire on the other. At what angle should the gun be aimed at the target ship in order to make a hit if the shot is fired at the instant when both vessels are on the straight line perpendicular to their course? The shell velocity v0 is assumed constant.

Homework Equations

The Attempt at a Solution

The answer in text is : theta = arccos( (v1 +v2)/v0)

The book doesn't have an explanation, and I am having trouble just being able to draw the situation. I don't understand what it means when the problem says " when both vessels are on the strait line perpendicular to their course."

 

[gneill]

The ships are traveling along straight paths. The paths are separated by some distance, thus they are traveling on parallel courses. At some instant in time they will be directly opposite each other, so that a line perpendicular to their paths can be drawn between them.

 

[gulfcoastfella]

Start by establishing your axes. For example, let the origin be located at one of the ships when it's path is perpendicular to the line of sight to the other ship, as given in the problem. Next, specify the x- and y-axes to be in the plane of the ocean, and the z-axis to be vertically pointed to the sky. Now specify the accelerations for each component and integrate from there.

 

[teddyayalew]

Thank you both for your replies here is the diagram I have drawn:

http://i42.tinypic.com/2lnzbxw.jpg

I am still having trouble understanding with the idea of finding the angle if this is a problem involving 3-dimensions. Because the cannon can be adjusted in the x and z direction. The question may seem silly to ask here, but I am studying mechanics on my own so I don't have teacher/tutor I can ask at my convenience so thank you for your replies.

 

[gulfcoastfella]

Are you expected to account for both the elevation angle and the horizontal angle?

 

[tms]

Are you expected to account for both the elevation angle and the horizontal angle?

Since the problem statement says the shell velocity is constant, either there is no gravity (in which case there is no need to elevate), or only the horizontal velocity matters (no air?).

 

[tms]

Start by establishing your axes. For example, let the origin be located at one of the ships when it's path is perpendicular to the line of sight to the other ship, as given in the problem. Next, specify the x- and y-axes to be in the plane of the ocean, and the z-axis to be vertically pointed to the sky. Now specify the accelerations for each component and integrate from there.

All speeds are given as constant, so there are no accelerations. The problem is simply to find out where the target will be when the shell gets there.

 

[tms]

The Attempt at a Solution

The answer in text is : theta = arccos( (v1 +v2)/v0)

The book doesn't have an explanation, and I am having trouble just being able to draw the situation. I don't understand what it means when the problem says " when both vessels are on the strait line perpendicular to their course."

Draw a diagram using the firing ship's frame of reference. The answer should jump out at you.

 

[teddyayalew]

@tms
thank you. I took your advise and drew the diagram and defined the velocities of the shell and boat to relative to boat one:
http://i43.tinypic.com/2u4q1ol.jpg

Then I solved for theta using the fact that the distance between the two boats when they are on the initial line can be represented in two ways. But the problem is when i solved for theta :http://i39.tinypic.com/dwumip.jpg
I did not get the correct answer. I know my algebra was correct so can you tell me if the way I defined the velocities was incorrect?

 

[teddyayalew]

If you cannot read the images I defined velocites like this:

Velocity(shell relative to boat one) = V(0) +V(1)
Velocity(boat two relative to boat one) = V(2) + (V1)

And from the given equations in the problem
V(0)= shell velocity relative to earth,
V(1) = velocity of boat 1 relative to earth
V(2) = velocity of boat 2 relative to earth

 

[tms]

If you cannot read the images I defined velocites like this:

Velocity(shell relative to boat one) = V(0) +V(1)

The question isn't entirely clear about this, but the shell's given velocity is relative to the gun. Assuming the correct answer you quoted is correct, that is.

Velocity(boat two relative to boat one) = V(2) + (V1)

Correct.

And from the given equations in the problem
V(0)= shell velocity relative to earth,
V(1) = velocity of boat 1 relative to earth
V(2) = velocity of boat 2 relative to earth

As I said, if you do everything in that boat's frame of reference the answer jumps out at you.

 

[teddyayalew]

Ok so instead of finding two ways to represent the initial distance of boat two from boat 1(the origin) I used boat 1 as the fram of reference but I still ended up getting
theta = arcsin( (v1 + v2)/v0)

Here is my diagram : http://i44.tinypic.com/ndk0fa.jpg

 

[tms]

The angle of bearing is usually measured from the bow, not the beam.

 

[teddyayalew]

Thank you for telling me that. Now I understand the problem!