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Another Two Dimensional motion question

  1. Sep 9, 2009 #1
    1. The problem statement, all variables and given/known data
    A second baseman tosses the ball to the first baseman, who catches it at the same level from which it was thrown. The throw is made with an initial speed of 17.0 m/s at an angle of 35.0° above the horizontal. (Neglect air resistance.)

    What is the horizontal component of the ball's velocity just before it is caught?

    3. The attempt at a solution
    i got the time but i keep getting the velocity wrong.
  2. jcsd
  3. Sep 9, 2009 #2
    You may be making this too complicated. You're given a vector at an angle of 35 degrees; try finding its components.
  4. Sep 9, 2009 #3
    what do you mean by the compents?
  5. Sep 9, 2009 #4
    The vectors in the horizontal and vertical directions that make up the vector you're given. The vector you're using has a magnitude of 17 m/s, at an angle of 35 degrees to the horizontal. Use trig functions to find the individual components.
  6. Sep 9, 2009 #5
    would you do this?
    sin35 = x/17 and solve for x
  7. Sep 9, 2009 #6
    No, the sine helps you solve for the vertical component. Try cosine.
  8. Sep 9, 2009 #7
    cos35 = x/17
    than would you use the final speed formula
    square root Vx^2+Vy^2
  9. Sep 9, 2009 #8
    You're making it too complicated~

    X is your horizontal component here. Solve 17cos35=x and you're done.
  10. Sep 9, 2009 #9
    really thats the velocity
  11. Sep 9, 2009 #10
    Yep. Your original vector is a velocity (speed in a direction), so therefore the component vectors are also velocities.
  12. Sep 9, 2009 #11
    i see, but i always thought that if you throw it x velocity in the begining that it would be the same velocity at the end
    and thanks :)
  13. Sep 9, 2009 #12
    But the ball's thrown at an angle. So it'll hit the first baseman's hit at the same speed that it left the second baseman's hand, yes. Assuming the second baseman's hand when the ball leaves it has the same elevation as the first baseman's mitt when he makes the catch. However, the component vectors are a different thing - since you're neglecting air resistance, the horizontal velocity remains the same throughout the throw, but the vertical velocity changes as the ball reaches the peak of its arc.

  14. Sep 9, 2009 #13
    lol. thanks for the help
  15. Sep 9, 2009 #14
    No problem :)
  16. Sep 9, 2009 #15
    ah, can you help me on this one too
    A swimmer runs horizontally off a diving board with a speed of 3.60 m/s and hits the water a horizontal distance of 1.81 m from the end of the board.

    How high above the water was the diving board?
  17. Sep 9, 2009 #16
    Alright, you know the swimmer's horizontal velocity is 3.6 m/s and he travels 1.81 m before hitting the water. You also know that acceleration due to gravity is 9.8 m/s, and his/her vertical velocity initially is 0.

    Figure out how long the swimmer takes to travel 1.81 m at 3.6 m/s. That's also how long it takes to get from initial elevation to water level - you've now got t.

    Then use the following formula:

    x = x0 + v0t + .5at2
  18. Sep 9, 2009 #17
    would you use thise equation to find t
    V^2 = Vo^2 + 2at
  19. Sep 9, 2009 #18
    opps.. i mean V^2 =Vo^2 + 2aX
  20. Sep 9, 2009 #19
    nope scratch that the first one
  21. Sep 9, 2009 #20
    No, you just need to use v = d/t. Use your horizontal information to calculate time.
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