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Need Help: Momentum and Energy

  1. Dec 5, 2009 #1
    1. The problem statement, all variables and given/known data
    A Hunter shoots a 500g arrow at a 2 kg bird perched on a tall tree growing on flat, level ground. The arrow is launched from ground level with a speed of 40 m/s at an angle of 30 degrees above the horizon. It is travelling horizontally when it embeds in the bird. How far from the base of the tree do the bird and aroow land

    2. Relevant equations
    Momentum and Energy Question

    3. The attempt at a solution
    V=40m/s*cos30 = 34.64
    P1=.5kg*34.64m/s= 17.32

    I have been supplied with the correct answer, but I must figure out how to get there (14m)
  2. jcsd
  3. Dec 5, 2009 #2
    Couple of different ways to tackle this. First you need h, and this can be easily solved by looking at the vertical component of the initial velocity, i.e. sin (30)* Vi. Once known, treat is as a simple ballistics problem with the new velocity you obtained by conserving mo.
  4. Dec 5, 2009 #3
    20 m in height
    Alright Ill try it again and if it diesn't work Ill post in like 5 minutes
  5. Dec 5, 2009 #4
    So i just tryed it again and got 17.8 so Im thinking one of my numbers I provided in my attempt posted above must be wrong
  6. Dec 5, 2009 #5
    What t did you get?
  7. Dec 5, 2009 #6
    I never got a t because Ek is mv^2 and Eg is mgh
    I didn't know t was relevant nor have I thought about how to get it from my givens
  8. Dec 5, 2009 #7
    Ok, as I said, there is more than one way to approach the problem. I verified the answer as 14 using t.
  9. Dec 5, 2009 #8
    Maybe if you show me the eqn you used to get the answer 17 I can help.
  10. Dec 5, 2009 #9
    I'll post in a sec
  11. Dec 5, 2009 #10
    So i got Vi with 40cos30 which gave me 34.64m/s
    i used momentum p1=p2 and did the following
    I got 6.93m/s
    from there I did Ek(arrow)+Eg(bird)=Ek(final)
    I tried using the bird for Eg but obviously that worked no better as my number came out as 23.5 but that is what i tried
  12. Dec 5, 2009 #11
    Ill be at work for the next 7 hours so I will be restricted to the work I can do but I will still check this site on my Iphone and can try stuff while it is slow at work
  13. Dec 5, 2009 #12
    But I still don't see where this leads you to x, the horizontal distance from the tree? The math looks good, what I did was to solve for t from the initial vertical velocity--you don't even need to compute h, but thats fine if you do. Here t is the time of ascent which is also the time of descent. You have computed the horizontal velocity after impact, you have t, the answer awaits.
  14. Dec 5, 2009 #13
    I know t?
  15. Dec 6, 2009 #14
    Need help on Plane Mirror

    Plane Mirror
    A person walks into a room that has, on opposite walls, two plane mirrors producing multiple images. Find the distances from the person to the first three images seen in the left-hand mirror when the person is 5ft from the mirror on the left wall and 10 ft from the mirror on the right wall.

    I know the first image is 10 ft because in the plane mirror, the image is as far behind the mirror as the object is in front. I don’t know how to figure out the second and the third images? I need you help please. Thank you.

  16. Dec 6, 2009 #15
    yo maybe we can finish my question first and then move on to yours
  17. Dec 6, 2009 #16
    Sure you can find t. The most straightforward is to get Vy from the initial data and divide by g. Say Vy=30m/s then t is about 3 seconds.
  18. Dec 6, 2009 #17
    Thanks man, you been a great help. I was stressing out about this before, but you really saved my @$$
  19. Dec 6, 2009 #18
    No problem.
  20. Dec 6, 2009 #19
    Sorry to post again but I read and made that last post at work.
    I am assuming you mean g is gravity(9.8) but that gives me 3.53 s. am I just supposed to use d=v*t to solve.
  21. Dec 6, 2009 #20
    Thats not right--the time is (sin 30* 40m/s)/9.8 should be about 2 seconds. In other words the vertical velocity is 20m/s.
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