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How to find final velocity?

  1. Sep 24, 2009 #1
    1. The problem statement, all variables and given/known data
    A baseball thrown at an angle of 65.0° above the horizontal strikes a building 18.0m away at a point 5.00m above the point from which it is thrown. Ignore air resistance.
    1- find the magnitude of the initial velocity of the ball.-found
    2- find the magnitude of the velocity of the baseball just before contact with the building.

    2. Relevant equations
    Vx=Vox+at
    X=Xo+Vox(t)+1/2at^2


    3. The attempt at a solution
    I successfully found question 1 which is the initial velocity of the baseball to be 16.3 m/s. I found this by using x=Xo+Vox(t)+1/2at^2 with the time I found plugged in. However, when it comes to the velocity of the ball just before it hits the building I get confused as to the difference or the comparision of the two velocities. How are they related?
     
  2. jcsd
  3. Sep 24, 2009 #2
    Do you know what it is asking for when it says the velocity of the ball just before it hits the building?

    (Hint: What variable?)
     
  4. Sep 24, 2009 #3
    Vxfinal?
     
  5. Sep 24, 2009 #4
    Yes, so lets see what variables you have:

    Vy Initial
    Displacement
    Acceleration (Gravitational Constant)
    Time

    And you are trying to solve for VxFinal.

    Which formula would you use?



    *Sorry, this may not be exactly what your asking... I don't know whether you want to know how to solve question 2, or as you stated, the relationship of the "2 velocities" [which I don't get what you mean by that.
     
  6. Sep 24, 2009 #5
    I would say Vy=Yo+Voy-1/2gt^2 but I'm not so sure that would solve for the velocity right before the ball hits the building. I am confused on what equation to use because none of them seem to serve a purpose for this question (number 2).
     
  7. Sep 24, 2009 #6
    Oh, so your confused on the wording. Yeah, I just had a similar question in my text that kinda threw me off for a split second, but the wording is theoretical.

    If you think of it that it said "final velocity" (IE: when the collision happened), the ball in theory would not be moving anymore (or it'd be bouncing off) and thus, there would be no velocity or it'd be in a different direction which is more advanced physics. When they say right before the collision, they basically mean like .0000000001 seconds before it collides and would bounce or come to a dead stop.

    Essentially, the difference that small is negligible in your calculations, so essentially, the formula you listed is solving for the same thing.

    If you understand calculus, it would be the limit as x (time) approaches the time of collision.
     
  8. Sep 24, 2009 #7
    So just solve for Vy? I'm not sure because that wouldn't take into account the Vox which i found above.
     
  9. Sep 24, 2009 #8
    Well, what do we know about the horizontal components of a projectile? What is special about them (as oppose to acceleration)?

    Hint: Solving for VyFinal would give you the y component of the final magnitude.
     
  10. Sep 24, 2009 #9
    Maybe if somebody could possibly work it out as far as I've gotten and could tell me if I did the first part right. It seems that I have lost my time figure aswell....:-/
     
  11. Sep 24, 2009 #10
    Oh sorry, I didn't look at your calculations closely. You can't do what you did because Horizontal Components of a projectile have uniform velocity (because they don't accelerate)

    So, horizontally the equation would be simply Vox = (d / t)

    Not any of the dynamics formula's with acceleration.

    So, the way you described I don't see how you got the velocity.

    For reference, what grade/college level are you taking this at so I know which formulas you know?
     
  12. Sep 24, 2009 #11
    It's college physics 1. And unfortunately the problem is much more complex than I would like it to be.
     
  13. Sep 24, 2009 #12
    Thanks for all of your help. I think i'm gonna forget about the last part of the question since I got the first part correct and the assignment is due tonight. It's also not very many points so I will just figure it out later. Thanks again!

    Kyle
     
  14. Sep 25, 2009 #13
    Well, if you end up going to go attempt it. This is how I would go about it.

    I would use the trajectory formula (since you know the end coordinates) to find out the only missing value Vi.

    Once you have Vi, you can then use Vox = dt (remember, there is no acceleration in horizontal motion) to find the time value.

    With the time value, you can use the
    Vfy^2 = Voy^2 + 2ad

    To solve for the final velocity of the y component. Then you have the x and y components of velocity which you can use to find the overall magnitude.
     
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