Momentum (Cannon fires at angle & recoils)

[CaptainSFS]

Homework Statement

A circus cannon, which has a mass M = 5000 kg, is tilted at q = 40°. When it shoots a projectile at v0 = 80 m/s with respect to the cannon, the cannon recoils along a horizontal track at vcannon = 1 m/s with respect to the ground.

1. At what angle to the horizontal does the projectile move with respect to the ground?

Homework Equations

The Attempt at a Solution

In order to solve this first question, I first found the horizontal velocity of the ball. 80cos(40) gives me about 61.xxx m/s. The cannon recoils in the opposite direction, so I add 1 m/s. I then use arccos(62.xxx/80) to find the new angle with respect to the ground. This is not the correct answer. I'm not sure what to do.

 

[Rake-MC]

An interesting question, according to the diagram, they want you to solve for $$\theta$$

Right?

Then please label the 40 degrees.

 

[CaptainSFS]

what do you mean "Then please label the 40 degrees"?

If you're implying that the answer is 40 degrees, I can assure with certainty that it isn't. I input these values into my homework online, and it returns with either an "OK" or "NO". I've tried 40 degrees, that was my first guess. =/

So, there's something else going on in the problem that I'm not understanding.

 

[Rake-MC]

The question is asking you to solve for theta in the diagram. Is this correct or not?

If this is correct. Then where is the 40 degrees in the diagram and what does it represent.
Your description states that q = 40 degrees. However q is not mentioned anywhere before or after that statement. What is q?

 

[CaptainSFS]

q is typically the letter they use to denote theta, since I cannot input a theta symbol into my answer field.

The field asks for "q_ground="

 

[CaptainSFS]

These are the hints they give me, I just don't understand at all what to do. I now believe though that the velocity I'm finding in the x-direction is incorrect, but I'm not sure why.

HELP: Because the cannon is moving, you must find the speed of the projectile with respect to the ground.
HELP: Recall from Lecture 3 (Relative Motion): The velocity of the projectile with respect to the ground equals the velocity of the projectile with respect to the cannon plus the velocity of the cannon with respect to the ground. Remember that velocity is a vector, so be careful of components and signs.