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Homework Help: Minimum velocity question

  1. Sep 25, 2005 #1
    !!!!!!!!!! Please Help !!!!!!!!!!!!!

    A cannon, located 60.0 m from the base of a vertical 25.0m-tall cliff, shoots a 15-kg shell at 43 degrees above the horizontal toward the cliff.

    I have determined:
    v=47.75 m/s
    v_0x= 34.92 m/s
    v_oy=32.56 m/s
    a_x=0
    a_y=-9.8 m/s^2
    x_o=0
    y_o=0
    x=60m
    y=25m
    alpha= 43 degrees

    part a: What must the minimum muzzle velocity be for the shell to clear the top of the cliff? answer=32.6 m/s


    Part b: The ground at the top of the cliff is level, with a constant elevation of 25.0 m above the cannon. Under the conditions of part (a), how far does the shell land past the edge of the cliff?

    How do I find part b
     
  2. jcsd
  3. Sep 25, 2005 #2
    Don't panic.
     
  4. Sep 25, 2005 #3
    Thanks for the help. I completely understand now.
     
  5. Sep 25, 2005 #4
    Figure out the velocity just as the shell clears the cliff. Then treat it as a new problem as if the edge of the cliff were your starting point. From there it's a simple 2d projectile motion problem on flat ground.
     
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