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Crossbones's Profile User Rating: -----

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Active Posts:
50 (0.03 per day)
Most Active In:
The Lounge (39 posts)
Joined:
14-January 11
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Last Active:
User is offline Sep 10 2011 02:50 PM
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Member Title:
Mercenary /Expendible Crew Member
Age:
Age Unknown
Birthday:
Birthday Unknown
Gender:
Male Male
Location:
Canada

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Posts I've Made

  1. In Topic: So Who Wants Something Made Custom?

    29 March 2011 - 07:10 AM

    HI Moati,

    Dibs on a custom vest. Email to follow tonight.

    Crossbones
  2. In Topic: True or False

    14 March 2011 - 07:52 AM

    True

    TNP is using the internet at work to view the forum....
  3. In Topic: True or False

    24 February 2011 - 07:55 PM

    False.

    TNP is eating pizza for dinner.
  4. In Topic: Physics Help (g.12)

    24 February 2011 - 06:44 PM

    I'm not sure if you are still trying to solve this but here is how I would do it.

    I will define the variable 't' = "the time spent traveling"

    As you may already know your problem describes a right angle triangle with the following sides:

    1. A = Long side adjacent to the 90░ = 500
    2. B = Short side adjacent to the 90░ = 3.7Ět (or 3.7 times 't')
    3. C = Hypotenuse = 10Ět

    Setting up the standard equation for a right angle triangle (Pythagoras Theorem) A^2 + B^2 = C^2 gives the following relation ship (the '^' hat symbol indicates "raised to the power of" so that each side is squared):

    Equation: 500^2 + (3.7Ět)^2 = (10Ět)^2
    Steps to solve: 250,000 + 13.69Ět^2 = 100Ět^2
    250,000 = 86.31Ět^2
    t^2 = 250,000/86.31 = 2896.54
    t= 53.8 seconds

    Solving for the angle is easy if you use the sine function. I'll define "theta" as the angle of traverse.

    sin(theta)= opposite/hypotenuse = (3.7Ět)/(10Ět)= 0.37

    so theta = asin(3.7) = 21.72░

    You could also use cosine or tangent but you have to multiply 't' into one of the velocities which is slightly longer math.

    I hope this helps.
  5. In Topic: True or False

    24 February 2011 - 06:16 PM

    True.

    TNP wears socks with sandals.

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