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Velocity Height Calculation check

  1. Oct 25, 2009 #1
    1. An object of mass 1.2kg is fired vertically in a barrel velocity 25n acting over a period of 0.25 seconds, calculate A) The velocity when leave the gun. B). Height reached. C). Time taken to return to initial height.



    2. a= f/m, V^2=U^2+2as



    3. A), A=F/M for acceleration, 25/1.2= 20.8
    Velocity = V0+AT = 0+20.8x0.25=5.2m/s

    B). Height reached = V^2=U^2 + 2AS, 0=5.2^2 + 2 (-9.81)s
    0= 27 - 19.62 x s, S = 27 / 19.62 = 1.38m

    C). Time, S = (T/2( (UxV), T= 1.38 x 2/5.2 = 0.5
    as require to arrive back = 2 x 0.5 = 1 Second

    any comments appreciated
     
  2. jcsd
  3. Oct 25, 2009 #2

    Andrew Mason

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    The velocity leaving the barrel is not correct.

    In A you are not taking into account the fact that it is fired vertically. Do a free body diagram showing the forces acting, including the 25 N applied force.

    AM
     
  4. Oct 27, 2009 #3
    Am I correct in understanding that
    V0+at
    Should be V0-Gt?
     
  5. Oct 27, 2009 #4

    Andrew Mason

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    No. Do a free body diagram. What forces act on the object? What is the net force? What is the acceleration? Use that value for "a" in the equation v = v0 + at.

    AM
     
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