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Projectile Motion ship gun

  1. Jun 22, 2009 #1
    1. The problem statement, all variables and given/known data
    An enemy ship is on the western side of a mountain island. The enemy ship can maneuver to within 2500 m of the 1800 m high mountain peak and can shoot projectiles with an initial speed of 250m/s. If the eastern shore line is horizontally 300m from the peak, what are the distances from the eastern shore at which a shop can be safe from the bombardment of the enemy ship?


    2. Relevant equations



    3. The attempt at a solution
    I got what ranges you could be farther away than or x>3577.55, but I can't find a way to get the x< part. I have y=Tanθx-gx²/(125000Cos²θ). And I made an inequality where y>1800. I don't think this is the right approach though. Also I am not sure if the ship can move back or not. If it can't please tell me and then I would know how to do this.
     
  2. jcsd
  3. Jun 23, 2009 #2

    tiny-tim

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    Hi blackboy! :smile:
    The shop moves but the ship doesn't. :wink:

    (if i'd written the question, the projectiles would have been sheep. :rolleyes:)
    Why do you still have θ in your equations?

    The x and y equations should have cosθ and sinθ, but then you should eliminate θ (squaring and adding usually works).

    Show us what you get. :smile:
     
  4. Jun 23, 2009 #3

    andrevdh

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    You could try this approach:

    The projectile needs to clear the peak. Consider the peak to be a point on the left-hand side of the "initial" parabola and see at what angle it should be launched to pass through it. Try and figure out what will happen if is launched at higher elevations, that is will it clear the peak? This will happen if the centre of the "initial" parabola is to the right of the peak.

    The closest that the shot can get is when the peak is a point on the right-hand side of the parabola isn't it?
     
    Last edited: Jun 23, 2009
  5. Jun 23, 2009 #4
    Ok so then can't I solve for θ in my quadratic if I make everything in terms of Tanθ and have that parabola going through the point 2500,1800?
     
  6. Jun 24, 2009 #5

    tiny-tim

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    el … im … in … ate !

    Yes, you could …

    but why bother when the question doesn't ask for θ …

    don't try to understand the little varmint :rolleyes:

    just eliminate it! :biggrin:

    (oh … and that'ld only give you the angle to hit the shop if it's on the edge of the cliff :wink:)
     
  7. Jun 24, 2009 #6

    andrevdh

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    Yes, that will give you the two launching angles and their subsequent ranges. The larger angle gives the smaller range since the projectile passes throught the point on its way down and the smaller angle gives the larger range since it passes the point on its way up.
     
  8. Jun 24, 2009 #7
    I got my answer with the solving for the angle way. x>3480 or x<270. Tiny Tim I don't understand your method, can you please explain?
     
  9. Jun 24, 2009 #8

    andrevdh

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    Your answers don't seem right to me. If I work backwards from them I get that the projectile will be at a height of 1262 m for your x = 3480 m (theta = 40,0 degrees) and 119 m for your x = 270 m (theta = 14,4 degrees) at the position of the peak?
     
  10. Jun 24, 2009 #9
    The angles I got were 50.1 and 75.6. The ranges were then respectively 6280 and 3070. Then subtracting 2800 I got 3480 and 270. Also can you tell me why you we want the parabola going through 2500,1800? I would imagine we could get closer to the eastern shore if we got a higher angle?
     
  11. Jun 25, 2009 #10

    andrevdh

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    Yes, the range would be less but the projectile will hit the mountain. That is the axis of the parabola is on the western side of the mountain. Launching at a higher elevation shifts it even further westwards causing the projectile to pass the peak at a lower height, and crashing into the mountain. Try it, calculate y(2500) for a slightly larger angle.
     
    Last edited: Jun 25, 2009
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