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Projectile motion -- launched 60° above the horizontal

  1. Jan 23, 2015 #1
    1. The problem statement, all variables and given/known data
    A projectile is launched at an initial speed of 30 m/s at an angle of 60° above the horizontal. Calculate the magnitude and direction of its velocity (a) 2s and (b) 5s after launch

    2. Relevant equations
    x=v*t in the x dimension
    x= x+vt+(1/2)at^2


    3. The attempt at a solution
    I tried going through with the x dimension to solve for acceleration but the acceleration in the y is 9.8.
     
  2. jcsd
  3. Jan 23, 2015 #2

    SteamKing

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    You need to be clear about which coordinate (x or y) represents the horizontal direction and which represents the vertical direction.

    If x represents the horizontal direction, why do you think the projectile is accelerating in that direction?
     
  4. Jan 24, 2015 #3
    Given that x is horizontal and y is vertical, the starting point is to break the launch velocity into its horizontal vector (x) ( which is deemed to be constant velocity )
    and its vertical vector (y) which comes under the influence of gravitational deceleration.
    Ive attached a sheet, which won't solve your problem but might give you a start.
     

    Attached Files:

  5. Jan 24, 2015 #4
    The velocity change will only take place in the y direction. Calculate the x component of velocity and the y component after 2 seconds and find their resultant.
     
  6. Jan 25, 2015 #5
    Yep thank you I was going in both.
     
  7. Jan 25, 2015 #6
    Thank you very much that helped Immensely!
     
  8. Jan 25, 2015 #7
    Than you very much this made it so much easier!
     
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