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Calculating the Vertical component of the Velocity

  1. Oct 25, 2014 #1
    1. The problem statement, all variables and given/known data
    A physics class researching projectile motion constructs
    a device that can launch a cricket ball. The launching
    device is designed so that the ball can be launched at
    ground level with an initial velocity of 28.0 m s–1 at an
    angle of 30.0 degrees to the horizontal.

    2. Relevant equations
    Calculate the vertical component of the velocity of the ball:
    a initially
    b after 1.00 s
    c after 2.00 s.
    3. The attempt at a solution
    Done A how do I do B and C??

    Thankyou
     
  2. jcsd
  3. Oct 25, 2014 #2

    Orodruin

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    Hi and welcome to Physics Forums!

    We require you to show some effort in working towards a solution before we provide help and guidance. What are your own thoughts on how to approach B and C?
     
  4. Oct 25, 2014 #3
    I thought about using the acceleration equations though I have not used these before with a degrees in the mix and was unsure if I needed to include this.
     
  5. Oct 25, 2014 #4

    Orodruin

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    So how do you think the ball is accelerated? Which component of the velocity is affected?
     
  6. Oct 25, 2014 #5
    The vertical component because horizontal does not use the acceleration equations
     
  7. Oct 25, 2014 #6

    Orodruin

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    Yes, so try applying that and show us what you get.
     
  8. Oct 25, 2014 #7
    so using the initial vertical component at the initial velocity in the equation v=u+at?
     
  9. Oct 25, 2014 #8

    Orodruin

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    Yes. As you said yourself, this is the only component of velocity that is influenced by the acceleration.
     
  10. Oct 25, 2014 #9
    Thankyou!!! Now I have it is seems easy! just wondering, why is the 9.8 ms-2 negative? is is because when it is moving up it is going in the opposite direction? If so does this mean that when is is moving downwards the 9.8ms-2 would be positive?
     
  11. Oct 25, 2014 #10

    Orodruin

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    This depends completely on how you have defined your coordinates. If you have defined the up direction as the positive y-direction, then the gravitational acceleration will be in the negative y-direction, i.e., down. If you did it the other way around and defined the down direction as positive, then you would have a positive gravitational acceleration, but the sign of the initial velocity in that direction would be changed as well.
     
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