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Conservation of Energy cannonball

  1. Feb 27, 2008 #1
    1. The problem statement, all variables and given/known data

    A 20.0 kg cannonball is fired from a cannon with a muzzle speed of 1050 m/s at an angle of 42.0° with the horizontal. A second ball is fired at an angle of 90.0°.

    (a) Use the isolated system model to find the maximum height reached by each ball.

    (b) What is the total mechanical energy of the ball-Earth system at the maximum height for each ball? Let y = 0 at the cannon.

    2. Relevant equations

    Ei = Ef

    K = (1/2)mv^2

    U = mgy

    3. The attempt at a solution

    I know that you can solve for h by setting these two equal, but I don't understand why that works. I need to understand how these relate. I have no idea what to do for b. Thank You.
     
  2. jcsd
  3. Feb 27, 2008 #2
    The total mechanical energy of the ball at any point is gonna be its kinetic energy + potential energy, and we're gonna say the ground has gravitational potential energy of 0

    So when either ball is first fired it's at a height of 0, so they have 0 initial potential energy, all they have is kinetic.

    At their maximum height their velocities are 0, so they have no kinetic energy, instead all they have is potential. Because of conservation of energy(and because we're ignoring air resistance and all that), ALL that initial kinetic energy was converted to potential, and though the problem doesn't deal with it, the ball is gonna fall and all that potential energy is gonna get turned back into kinetic, and the total amount never changes

    Also keep in mind for the first ball that it's fired at an angle, and you're only concerned with height in this problem, so you need to find the velocity in the vertical direction
     
  4. Feb 27, 2008 #3
    How do you find velocity in the vertical direction?
     
  5. Feb 27, 2008 #4
    I see on the 2nd ball that is fired the potential energy is 1.10e7 which is just mgy. How is the 1st ball calculated? I guess that's where I am having the most problem with that angle. Is it still just mgy?
     
  6. Feb 27, 2008 #5
    average velocity:
    LaTeX graphic is being generated. Reload this page in a moment.
     
  7. Feb 27, 2008 #6
    yes your on the right tracks
     
  8. Feb 27, 2008 #7
    So for the 90 degree ball you just took its initial velocity(which was all vertical since it was at 90 degrees) and found the initial kinetic energy, and knew that was how much potential energy it'd have at the peak

    For the other ball you do the exact same thing, but it'll have a different initial velocity in the y direction. You're on an energy unit so you've done projectile motion, which means you've found initial vertical(and horizontal though you don't need it here)velocities given initial velocity and an angle about 44594895709 times. Don't overthink it
     
  9. Feb 27, 2008 #8
    Ok, so I would use mg (Vsin(42))?
     
  10. Feb 27, 2008 #9
    Is this correct?
     
  11. Feb 27, 2008 #10
    Why would you want m*g*Vyi?
     
  12. Feb 27, 2008 #11
    so I should be looking for mg*Vyxi?
     
  13. Feb 27, 2008 #12
    Well what good is mgVanything to you? That's almost your equation for potential energy

    You CAN use that velocity to find initial kinetic energy though. Then you can use potential energy to find distance
     
  14. Feb 27, 2008 #13
    Well I am looking for the potential energy of the 1st ball on part b which has the 42 degree angle. All of the energy is still potential right?
     
  15. Feb 27, 2008 #14
    I'm still lost?
     
  16. Feb 27, 2008 #15
    Well eventually, but initially all it has is kinetic
     
  17. Feb 27, 2008 #16
    Ok, then what do I need to solve the problem? Does the angle matter?
     
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