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How to calculate velocity with target 12 m away vertically and 6 m horizontally?

  1. Feb 13, 2012 #1
    I used an online simulation for this lab. The target is 12 m away horizontally and 6 m vertically. I was to find the velocity needed to hit the target.Since it's an online simulation, I played around with it and got 14.3 m/s. The formula I plan on using is:

    Vx=range/time
    Vy=√(range*10-(range/time)²)
    V=√(Vx)+Vy

    My problem is, how do I calculate the range and time when the target is hit? Or do I use the final range and time in the equation?

    The range and time of the first black X is: 9.19 m and 1 s.
    The final range and time is 21.6 m and 2.4 s.
     
  2. jcsd
  3. Feb 13, 2012 #2
    I was given these equations and was told to solve for V:

    t = 12/Vx
    Vy = 6/t + (0.5)(9.81)t
    Vx = Vcos(50)
    Vy = Vsin(50)

    And this is what I got:

    t = 12/Vx
    2.6=12/Vx
    Vx=28.8

    Vy = 6/t + (0.5)(9.81)t
    Vy=6/2.6+(0.5)(9.81)(2.6)
    Vy=14.272

    Vx = Vcos(50)
    28.8/cos(50)=V
    V=29.8456

    Vy = Vsin(50)
    14.272/sin(50)=V
    V= -54.395

    The velocity from Vy = 6/t + (0.5)(9.81)t is the same from the simulation, but I think I might have messed up in general. I know V from both equations should be the same, but they're obviously not. Any explanation on what I did wrong and how to fix it?
     
  4. Feb 13, 2012 #3
    What angle with the horizontal is specified? I see the number 50 in your work but don't see where it came from. Without an angle specified, the problem does not have a unique solution.
     
  5. Feb 13, 2012 #4
    The angle of the cannon is 50°.
     
  6. Feb 13, 2012 #5
    Take your equations

    t = 12/Vx
    Vy = 6/t + (0.5)(9.81)t
    Vx = Vcos(50)
    Vy = Vsin(50)

    and rewrite them with V as one of the unknowns. The other unknown is t. Eliminate t and solve for V. Eliminate the Vx and Vy variables because you know them in terms of V which is what you seek.
     
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