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Relative motion of a cannonball/submarine

  1. Oct 4, 2014 #1
    1. The problem statement, all variables and given/known data
    1.)
    A projectile is fired from a submarine traveling horizontally at 20 m/s with respect to the water as shown in the figure below. According to an observer on the submarine, the projectile is fired at 45° with an initial velocity of 60 m/s. After firing the projectile, the submarine continues to travel at 20 m/s.

    According to an observer watching from a boat that is stationary with respect to the water, what will be the angle θ that the projectile makes with respect to the horizontal when it is launched?


    2.)A tugboat captain also sees the submarine fire the projectile, but to him is looks like the projectile is moving straight up (i.e. 90° above the horizontal). What is the velocity of the tugboat relative to the water?

    2. Relevant equations
    V[a][/b]=V[a][/c]+V[c][/b]

    3. The attempt at a solution
    I have no idea how to even start this one. The angle is really throwing me off. Any and all help to understand would be appreciated.
     

    Attached Files:

  2. jcsd
  3. Oct 4, 2014 #2

    BiGyElLoWhAt

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    You need to look at the relative velocities here. What would be the relative x component of the velocity of the projectile with respect to me be if, when I look down, I see it coming straight up. I know that's more part b, but I think it's easier to look at that one. You use the same concepts for both sections.
     
  4. Oct 4, 2014 #3
    ok... i think i see what you're saying. the x component of velocity if it looked like it was going straight up would be 0? because its 60cos(90)?
     
  5. Oct 4, 2014 #4

    BiGyElLoWhAt

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    right. 0. So there's b. Can you apply that logic in the same way to solve for the angle in a? How would YOU set it up?
     
  6. Oct 4, 2014 #5
    its telling me the answer for b is 22.4 m/s in the opposite direction of the submarine though
     
  7. Oct 5, 2014 #6

    Orodruin

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    You really should do part 1 first. The motion in the opposite direction should not be surprising in part 2, since the projectile is fired backwards.

    To start with, what are the horizontal and vertical components of the projectile velocity in the submarine frame? How are those going to be related to the components in the stationary frame?
     
  8. Oct 5, 2014 #7
    horizontal component is 60cos 45 and vertical is 60sin 45. Won't those stay the same in a stationary frame?
     
  9. Oct 5, 2014 #8

    BiGyElLoWhAt

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    Are you sure you should be using 60 for the amplitude? You want the velocity of the cannonball with respect to the submarine (per Oridruins question)
    For a, what do you know about the angle of a trajectory with respect to it's velocity?

    If I give you a velocity, say <a,b> and I asked you what the angle of it's motion was, what would you do?
    (this is the last step of solving a, but I think it's often a good idea to see what your ultimate goal is in order to start the problem)
     
  10. Oct 5, 2014 #9
    I would do the tan inverse of that if it was the horizontal and vertical components. I can tell the cannon ball would be affected differently in the sub's reference frame because the submarine is going the opposite direction but i'm not sure how to quantify that
     
  11. Oct 5, 2014 #10

    BiGyElLoWhAt

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    What's the submarines velocity in components? What's the cannonballs velocity RELATIVE TO THE SUBMARINE'S VELOCITY in components?

    In other words ##\vec{v}_{cannonball}= \vec{F}(\vec{v}_{submarine})## where F is some function of the submarines velocity. Can you look at the situation and express the RHS of the equation? (This might look a little more challenging, but don't over think it, it's not too bad)
     
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