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Finding the magnitude of velocity vector

  1. Jun 9, 2017 #1
    1. The problem statement, all variables and given/known data
    A cricketer throws a 147g ball by exerting a force of 7.5N for 0.84s. If the launch angle is 39 degrees from horizontal, calculate the range of the ball.


    2. Relevant equations


    3. The attempt at a solution
    I have the range equation and I just need to find the magnitude of the velocity vector but I'm not sure how I'd find that with the given quantities?
     
  2. jcsd
  3. Jun 9, 2017 #2
    What is the answer given in solutions section?
     
  4. Jun 9, 2017 #3

    SammyS

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    Can you determine the acceleration ?
     
  5. Jun 9, 2017 #4
    I thought it was a=f/m = 51.02m/s^2 and then v=at for 42.86 ms^-1 I was going to use a particular formula for range

    The solution in the back of the mock exam says f=ma rearranged for m x deltav/deltat

    Which is then rearranged for deltav = Fdeltat / m = 42.8 ms

    Trigonometry to then find the Vx and Vy components which can then be plugged into t = v-u/a giving us a time value. This completes the vertical plane suvat.

    Horizontal 'range' was simply given as displacement s=vt: 42.8(cos39) x 6.78 = 225.52m

    upload_2017-6-9_20-58-45.png
    When would this equation for range be applicable? It's defined as 'the displacement in the x axis when the y axis displacement is 0.' I imagine there is still some displacement for this scenario even though it doesn't end in the y axis? Making this equation irrelevant?
     
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