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Kinetic motion

  1. Jun 10, 2007 #1
    1. The problem statement, all variables and given/known data
    A bottle is thrown with an initial velocity of 4 m/s at 45 degree angle from the horizon. Find its final horizontal and vertical velocities before striking the ocean.


    2. Relevant equations
    change x= innitial velocity (t) + a(t squared)
    final velocity= innitial velocity + at
    d= average velocity (t)


    3. The attempt at a solution

    I don't know how to solve this equation because all of the formulas include time and the problem doesn't give time.
     
  2. jcsd
  3. Jun 10, 2007 #2
    I think you are missing how far below the ocean is below from where the bottle is thrown. Also, you know that the x velocity will be constant because there is no acceleration in that direction. So you will simply have to figure out the y component.
     
  4. Jun 10, 2007 #3
    it says from the horizon, so I guess the innitial and final heights are the same. I have the answer btw, I jsut don't know how Kaplan got it. here's the answer...2.8 m/s in both the x an y directions.
     
  5. Jun 10, 2007 #4
    You might want to take a look at this. Click the next button at that site and review how these kinematic equations are used.

    To get you started, consider:

    You don't need to use any of the kinematic equations :surprised

    You know the initial velocity. It's a vector, right? You know its magnitude and direction.
    What's the magnitude of its vertical component? (Hint: this where the 2.8 came from. Think trig).
    How about its horizontal component? Will the horizontal component change?
    How about the vertical component?

    Will the bottle's vertical motion not come to zero at some maximum height and then accelerate back down toward the ocean?

    You didn't specify any difference between the height above the ocean when the bottle left the thrower's hand and the ocean level itself. Therefore, if the bottle decelerates to zero from the intial vertical velocity, then it accelerates back down over the same distance that it went up, what must be the final vertical velocity?
     
    Last edited: Jun 10, 2007
  6. Jun 10, 2007 #5
    aha, I see the 2.8 now. I had calculated the sq. rt. of 2 / 2 and the only thing I didn't do was multiply that by four. thanks
     
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