1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Skeet shooting

  1. Nov 22, 2009 #1
    1. The problem statement, all variables and given/known data

    A 0.25 - kg skeet is fired at an angle of 300 to the horizontal with a speed of 30 m/s. When it reaches the maximum height, it is hit from below by a 15-g pellet traveling vertically upward at a speed of 200 m/s. The pellet is embedded in the skeet.


    a) How much higher does the skeet go up?

    b) How much extra distance x, does the skeet travel because of the collision?



    2. Relevant equations



    3. The attempt at a solution

    ms = .25. vs = 30... mp = 15000. vp = 200..Q=30 degrees

    msvs + mpvp = (ms + mp)v...15000(200) = (.25 + 15000)v
    v = 200 = vy'

    vx = 30cos30 = vx'=25.98

    t = x/30cos30 = x/25.98

    y = yo+vy't - .5gt2 = yo + 200(x/25.98) - .5g(x/25.98)2 = yo + 8.25x - x2/.01 ...x = .077 is this correct so far
     
  2. jcsd
  3. Nov 22, 2009 #2

    Delphi51

    User Avatar
    Homework Helper

    I don't follow your work for the vertical velocity after the collision.
    momentum before = momentum after collision in vertical direction
    mv = mv
    .015*200 = (.015+.25)v
    solve for v.
     
  4. Nov 23, 2009 #3
    the pellet is 15g = 15000 kg
     
  5. Nov 23, 2009 #4
    Ask yourself if that makes sense. A bullet hitting a skeet will not make the skeet travel at the same velocity as the bullet. It will increase its velocity but not by that much. Looking at the mass of the pellet, does it make sense that it's mass is 15000kg? No, a gram is 1/1000 of a kilogram, not 1000kgs.
     
  6. Nov 23, 2009 #5
    sorry i always do that

    ms = .25. vs = 30... mp = .015. vp = 200..Q=30 degrees

    msvs + mpvp = (ms + mp)v... .015(200) = (.25 + .015)v
    v = 11.32 = vy'

    vx = 30cos30 = vx'=25.98

    t = x/30cos30 = x/25.98

    y = yo+vy't - .5gt2 = yo + 11.32(x/25.98) - .5g(x/25.98)2 = yo + .44x - (x^2)/.01 ...
     
  7. Nov 23, 2009 #6

    Delphi51

    User Avatar
    Homework Helper

    v = 11.3 is correct. I don't understand why your last line y= formula has x's in it. The horizontal motion does not affect the vertical motion.
     
  8. Nov 23, 2009 #7
    x = vxt.....t = x/vx

    isnt that how u typical solve these problems....
     
  9. Nov 23, 2009 #8
    You don't need to do that for this problem. You can use the quadratic formula to find t in your y = equation. Since the question asks for how much higher and farther the skeet travels I would suggest first finding the height and distance when it isn't hit, then finding height and distance when it is hit.
     
  10. Nov 27, 2009 #9
    ok so y = yo + vyt - .5gt2 = yo + 11.32t - .5(9.8)t2....ymax @ t = 1.16

    so y = yo+6.54

    x = xo + 25.98(2.31) = xo + 60.01

    is this correct, do i need to find the initials before i can solve the quadratic or can i do as i have done and say it is zero and then add it back in after i find t
     
  11. Nov 27, 2009 #10

    Delphi51

    User Avatar
    Homework Helper

    The 6.54 looks good for part (a).
    The 1.16 seconds must be the time of maximum height. I don't know if that will help you find (b). For (b) you are starting over with a new trajectory problem with Vy = 11.3. You'll have to find the x and y coordinates at that point where the pellet hits it and the Vx left from the initial shot. Then you can do the horizontal and vertical formulas to find the horizontal distance.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Skeet shooting
  1. Shooting an Arrow (Replies: 8)

  2. Skeet Shooting (Replies: 4)

  3. Monkey Shoot (Replies: 11)

Loading...