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Projectile motion and time

  1. Feb 12, 2008 #1
    [SOLVED] Projectile motion and time

    1. The problem statement, all variables and given/known data

    A hot air balloon rises from the ground with a velocity of (2.00 m/s) in the y direction. A champagne bottle is opened to celebrate takeoff, expelling the cork horizontally with a velocity of (5.00 m/s) in the x direction, relative to the balloon. When opened, the bottle is 6.00 m above the ground.

    How long does the cork remain in the air?


    2. Relevant equations

    There are other parts of this problem that I've already solved. For example, I've found that the initial velocity and direction of the cork, as seen by an observer on the ground, is 5.39 m/s at 21.8 degrees above horizontal (w/ respect to the balloon) with these equations:
    V= [(5.0)^2+(2.0)^2]^1/2

    theta= arctan (2.0/5.0)

    I've also found that the maximum height above the ground attained by the cork is 6.204 meters by finding:

    h= [(5.39^2)sin^2(21.8)]/2(9.81)

    and adding it to the initial height of 6.00 m.

    3. The attempt at a solution

    I know that the equation to find the time that a projectile is in the air is as follows:

    t= [2Vo*sin(theta)]/g

    but I know that that only accounts for the time that it will take the cork to reach a height of 6 meters. What I don't know is how to account for the rest of the time.
     
  2. jcsd
  3. Feb 12, 2008 #2
    Forget about the corks velocity in the x direction, all you need to know is the vertical. The question tells you that the cork is moving 2 m/s upward and starts off 6m above the ground.

    From there you can use constant acceleration formula to determine what the maximum height of the cork is and how long it takes to get there, and then how long it takes to fall to the ground again.
     
  4. Feb 12, 2008 #3

    Hootenanny

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    Welcome to PF,

    I'm not checking your math, but your method looks good to me. For the time of flight, forget your time equation there and consider one of the general kinematic equations (found here). You know the initial displacement, the initial velocity as well as the final displacement, so which equation do you think you should use?
     
  5. Feb 12, 2008 #4
    I guess it would be the displacement and time equation with uniform acceleration, just in the y dimension instead of the x. I'll get back with you after I do the math to see if I've got this figured out.

    Thanks so far, guys.
     
  6. Feb 12, 2008 #5
    Finished

    Awesome. It was that equation after all. I actually tried it before I posted the first time, but I think my math was off somewhere when I used the quadratic formula to solve for t. Then I just ended up overcomplicating things for myself. Sometimes you're your own worst enemy, y'know?

    Thanks again for the help. I really appreciate it.

    -Anne
     
  7. Feb 12, 2008 #6

    Hootenanny

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    I know the feeling :rolleyes:
    A pleasure :smile:
     
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