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Physics hw help

  1. Feb 28, 2010 #1
    a cannon located 60m from the base of a vertical 25m tall cliff, shoots a 15kg shall at 43 degrees above the horizontal toward the cliff (a) what must the minimum muzzle velocity be for the shell to clear the top of the cliff? (b) the ground at the top of the cliff is level, with a constant elevation of 25m above the cannon. Under the conditions of part (a) how far does the shell land past the edge of the cliff?
    I got part (a) to be 32.63 m?s but i dont know where to start for part b.
     
  2. jcsd
  3. Feb 28, 2010 #2
    When the shell lands, the y distance is 0. You can find the time for this and how far out the shell lands using the same motion equations you used for part a.
     
  4. Feb 28, 2010 #3
    but arent i finding the where it would land on the y axis?
     
  5. Mar 1, 2010 #4
    The y axis is height, so you need to know at what point in time the height is 0 (that is, the object is on the ground). Since the equation is a quadratic, you will have two answers. 1 will be 0 (since it starts from ground height) and the other will be the time at which it hits the ground on the other side of the parabola. Use that to find the x distance.
     
  6. Mar 1, 2010 #5
    i am getting the answer to be 21.9264m is that right?
     
  7. Mar 1, 2010 #6
    Not what I got, what did you get for the time?
     
  8. Mar 1, 2010 #7
    t= 2.51 sec
     
  9. Mar 1, 2010 #8
    Hmm.

    [tex] d_f = \frac{1}{2}at^2 + v_it [/tex]?
     
  10. Mar 1, 2010 #9
    now i have to answers for t 2.04sec and 2.5sec
     
  11. Mar 1, 2010 #10
    Wait, 2.51 is the time that the shell clears the cliff, we want where it lands. Set the y equation equal to 0 and then plug that time (and the same velocity from part a) into the x equation.
     
  12. Mar 1, 2010 #11
    what y and x equations?
     
  13. Mar 1, 2010 #12
    Y: [tex]d_y = \frac{1}{2}at^2 + v_it [/tex]
    X: [tex] d_x = x_it [/tex]

    You're doing the same thing you did to make sure the shell cleared the cliff and find the velocity for that (you didn't know time or velocity) except now you don't know time or distance.
     
  14. Mar 1, 2010 #13
    ok so t= 5.77 sec
     
  15. Mar 1, 2010 #14
    I'm sorry I'm still not getting that and I'm 99% sure my work is correct. Why don't you show me your work.
     
  16. Mar 1, 2010 #15
    0=-25=32.64t-4.905t^2
    t= -.88s t=5.77s
     
  17. Mar 1, 2010 #16
    OK, thanks.

    First off, I assume the -25 is for the height of the cliff which isn't needed. We already proved that the shell would clear the cliff in part a, so start back at the cannon from ground level.

    Secondly, the shell isn't shot horizontally, as your equation as it. It was shot 43º above horizontal, which affects the velocity.
     
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