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Basic vector questions

  1. Jul 25, 2014 #1
    This is problem very basic for all of you but I am struggling to grasp this.

    Heres the problem
    A cricketer is running due north with a cricket ball at 10 m/s and throws the ball due west at 24 m/s. Find the direction of the ball and its speed.

    Please give me a details explanation as I am completely 100% lost with this.
  2. jcsd
  3. Jul 25, 2014 #2
    You can define two vectors given your data, say ##\vec u## and ##\vec v##:
    1. Can you find ##\vec u+\vec v## ?
    2. What does the vector ##\vec u+\vec v## represent?
  4. Jul 25, 2014 #3
    Worked out the speed 26 m/s Answer sheet tells me 67 degrees 23 min west of north??? How does it arrive at that??
  5. Jul 25, 2014 #4


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    Homework Helper

    Draw the triangle that is made up of the (final) velocity vector and it's Northern and Western components.

    Can you use the known information to find the direction?
  6. Jul 25, 2014 #5


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    Science Advisor

    Did you understand what Hakim Philo said? The cricketer is running N at 10 m/s. You can take his velocity vector as a vector with length 10 pointing N. The ball, which was moving along with the cricketer, is thrown W at 24 m/s. You can take its velocity vector, relative to the cricketer, to be a vector with length 24 pointing W. Since N and W are at right angles, their vector sum is the hypotenuse of a right triangle with legs of length 10 and 24. You say you got 26 m/s for the speed of the ball, relative to the ground. I presume you got that using the Pythagorean theorem. The angle can be found by using the fact that if a right triangle has "opposite side" of length a and "near side" of length b, then the angle is given by [itex]tan(\theta)= a/b[/itex] so that [itex]\theta= arctan(a/b)[/itex].
  7. Jul 25, 2014 #6
    ahh got it. Thanks for the help. Turns out I drew the triangle upside down so of course I was never going to get the degrees correct. :)
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