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Energy - Cannon Ball and Conservation of Energy

  1. Apr 3, 2013 #1
    Energy -- Cannon Ball and Conservation of Energy

    1. The problem statement, all variables and given/known data
    A cannon tilted up at a 30.0 angle fires a cannon ball at 80m/s from atop a 10m -high fortress wall. What is the ball's impact speed on the ground below?


    2. Relevant equations
    Conservation of Energy

    Ugi+KEi= KEf

    *please note that the ground is the zero for potential energy *


    3. The attempt at a solution

    Ugi+KEi= KEf

    mgh+.5mVi^2= .5mVf^2 (m cancels out)

    gh+ .5Vo^2= .5V1^2

    V1= Square root of {(gh+.5Vo^2)/.5}

    81.22m/s

    Is this right? I expected the velocity to be much greater since the object was in freefall.
     
    Last edited: Apr 3, 2013
  2. jcsd
  3. Apr 3, 2013 #2
    Try splitting the velocity into it's x and y components, and then put them into the energy equations
     
  4. Apr 3, 2013 #3
    You did everything correctly. The contribution of the height is very small, because the potential energy at 10 m is insignificant in comparison with the kinetic energy at 80 m/s. You may want to find out what height a body must fall from to attain 80 m/s, you will see it is much higher than 10 m.
     
  5. Apr 4, 2013 #4
    I think there is a missing data which is the height of the ball after it was shot (the point where K.E=0), because it stated that the cannon is tilted up 30 degrees and the result should be about projectile motion. The height for P.E should be more than 10m and at the end the result of the final velocity will be bigger when it reaches 0 m.

    Note: use the conservation of energy in projectile motion
     
  6. Apr 4, 2013 #5
    I think there is a missing data which is the height of the ball after it was shot (the point where K.E=0), because it stated that the cannon is tilted up 30 degrees and the result should be about projectile motion. The height for P.E should be more than 10m and at the end the result of the final velocity will be bigger when it reaches 0 m.

    Note: use the conservation of energy in projectile motion
     
  7. Apr 4, 2013 #6
    You may want to review your understanding of "conservation of energy".

    Indeed.
     
  8. Apr 7, 2013 #7
    Sorry for the late response you guys. First off, I would like to thank each and everyone of you for taking out the time to help me. As it turns out, the answer that I had was correct. Again, sorry for the late response, and thank you for the assistance.
     
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