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Motion in Two Dimensions (PLEASE HELP!)

  1. Sep 13, 2010 #1
    1. The problem statement, all variables and given/known data
    An artillery shell is fired with initial velocity of 300 m/s at 55 degrees above the horizontal. TO clear an avalanche, it explodes on a mountainside 42 secs after firing. What are the x- and y- coordinates of the shell where it explodes, relative to its firing point?


    2. Relevant equations
    Vox = Vo (cos [tex]\theta[/tex])
    Voy = Vo (sin [tex]\theta[/tex])

    These could be the wrong equations. Please solve. THANKS!
     
  2. jcsd
  3. Sep 13, 2010 #2

    cepheid

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    Welcome to PF,

    Those equations are correct, but insufficient to solve the problem. You need to know where the projectile will be at t = 42 s. To know that, you need to know the x and y positions of the projectile vs. time. You should have encountered kinematics equations that tell you that.

    NO. We don't do your homework for you here. That would NOT help you in any way whatsoever. We already know how to solve these problems. It's your turn to learn, and if you don't learn how to do it because we did it for you instead, then you'll be in trouble when the test comes around. If that argument doesn't convince you, then I should point out that by not showing what attempt you have made so far, you are in violation of PF rules, and nobody will help you until you demonstrate that you have made a decent effort yourself, as the rules demand.
     
    Last edited: Sep 13, 2010
  4. Sep 13, 2010 #3
    I apologize, I thought i posted my attempts.

    I used the previous equations to give me these solutions:
    Vox = +1991.41 m/s Voy = -299.93

    Then, I used these equations:

    [tex]\Delta[/tex]y = (-299.93)(42 s) + 1/2 (9.8)(42)2
    [tex]\Delta[/tex]y = -3953.5
    [tex]\Delta[/tex]x = (1991.4)(42)
    [tex]\Delta[/tex]x = 83638.8

    Let me know where I went wrong, if I did.
     
  5. Sep 13, 2010 #4

    cepheid

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    Without looking at anything else, it's clear that this can't be right. It doesn't make sense that the x-component of the velocity is larger than the total velocity. Can you post the trig formulas you used to calculate the velocity components?

    EDIT: Nevermind, you posted them in your original post. Okay, well something doesn't make sense, because the cosine and sine of an angle should always be less than or equal to 1, and there's no way you can get 1991 m/s by multiplying 300 m/s by some number that is less than 1. What did you get for cos(55) and sin(55)?

    EDIT 2: Oh yeah, and the rest of what you did looks fine, so once you get the velocities right, your answers for x and y should be correct.
     
    Last edited: Sep 13, 2010
  6. Sep 13, 2010 #5
    Ok, I fixed the x component:
    The new Vox = 6.64 m/s

    By using this new value in the new equation:
    [tex]\Delta[/tex]x = (6.64)(42)
    [tex]\Delta[/tex]x = 278.88

    The [tex]\Delta[/tex]y should be correct.

    What do you think?
     
  7. Sep 13, 2010 #6

    cepheid

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    That seemed way too small to me, so I looked into it, and you calculated your cosine and sine wrong. The problem is that your calculator is set to interpret angles as being measured in radians rather than in degrees. Therefore, you calculated the cosine of 55 radians instead of 55 degrees.

    You always have to be careful about checking what units the calculator is using for angles when you use the trigonometric functions.

    Also, whenever you calculate an answer, stop and think whether it makes sense or not. In this case it clearly doesn't. If you sketched the right-angled triangle formed by the velocity vector and its components, with the 55 degree angle in it, you'd be able to see roughly how V0y and V0x (the two sides) compared in size to V0 (the hypotenuse). The two side lengths are a substantial fraction of the hypotenuse length -- just by looking at it, you can see that the horizontal side clearly isn't a factor of 6/300 smaller than the hypotenuse. These sorts of tips and tricks for identifying outrageous/nonsensical results will help you catch the more obvious mistakes.
     
  8. Sep 13, 2010 #7
    I apologize. I don't know why I keep doing that.

    Well after switching to degrees:

    [tex]\Delta[/tex]x = 7227.1
    [tex]\Delta[/tex]y = 18965.1

    EDIT 1: If you are wondering, my new Vox= 172.073 and Voy= 245.75
     
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