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Projectile Motion

  1. Nov 4, 2007 #1
    A rocket-powered hockey puck has a thrust of 1.20N and a total mass of 1.50kg. It is released from rest on a frictionless table, 3.20m from the edge of a 3.60m p. The front of the rocket is pointed directly toward the edge.


    I have found that the acceleration of the hockey puck in the x-direction is 0.8m/s/s because of the F_thrust/m_puck = a_x thing that was found in Newton's Second Law representation of the forces in the x-direction. I also found that the time between the moment when the puck falls off of the table to when it hits the ground, by virtue of the fact that the puck falls under the influence of gravity only, is 0.857s.

    [tex]\Sigma[/tex](F_x) = F_thrust = m_puck * a_x
    [tex]\Sigma[/tex](F_y) = n - w = m_puck * a_y

    Now, with the picture and these things here, I have a little bit of trouble find the correct kinematics equations to put these pieces of information into a solution.
     
  2. jcsd
  3. Nov 4, 2007 #2

    rl.bhat

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    Using v^2 = u^2 + 2as find the velocity with which the puck leaves the table. This velocity remains the in the horizontal direction. Vertical velocity of the puck increases as it falls under gravity. Find the time it takes to reach the ground. The product of this time and horizontal velocity gives the distance of point from the table where the puck hits the ground. By the way how did you get t = 0.857s?
     
  4. Nov 4, 2007 #3
    What is u?
     
  5. Nov 4, 2007 #4
    I do not think I have ever seen this formula. Could you explain exactly what the symbols represent, I mean except for a, which I assume to represent acceleration.
     
  6. Nov 4, 2007 #5
    V is final velocity, u is initial velocity.

    You have to first find the velocity right when the puck leaves the table. then the horizontal velocity will be constant and Vy will be subject to gravity(but will be initially 0 at the time of leaving the table).
     
  7. Nov 4, 2007 #6
    What is s?
     
  8. Nov 4, 2007 #7
    Nevermind.
     
  9. Nov 4, 2007 #8
    So, the final velocity in the y-direction is 8.4m/s. Can anyone confirm this for me?
     
  10. Nov 4, 2007 #9

    rl.bhat

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    That is correct. (put these pieces of information into a solution.)But actually what is required in the problem?
     
  11. Nov 4, 2007 #10
    To find the distance from the table that the puck lands after it falls off of the table.
     
  12. Nov 4, 2007 #11

    rl.bhat

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    You know the final velocity in the y-direction. Using this velue find the time it takes to reach the ground. Multifly this time with the velocity of the puck when it leaves the table. You can find it using v^2 = u^2 + 2as Here u = 0 , a = 0.8 and s = 3.2m
     
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