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Hitting a moving object?

  1. Mar 17, 2012 #1
    1. The problem statement, all variables and given/known data
    You go to an archery competition and decide to compete in the feather drop event. In this event, the target is dropped to the ground at the instant that you release the arrow from your bow. You are placed 30 meters from the target. Assuming the bull’s-eye is a circle the size of a silver dollar where should you aim your arrow?


    2. Relevant equations

    Not sure where to begin, or if any equations are involved.

    3. The attempt at a solution
    I assume you need to aim under
     
  2. jcsd
  3. Mar 17, 2012 #2
    Hmmm.
    Its a simple kinematics problem.
    You have to release the arrow in such a way that it collides with the falling body.

    In terms of equations it means the at a certain time t, both the coordinates x,y of the arrow and body should br same so that they can collide.

    The answer you will get is really interesting.
     
    Last edited: Mar 17, 2012
  4. Mar 17, 2012 #3
    Sorry that I don't understand completely, its just that I won't take a physics course till next year. Is there a specific equation that I need to use? I looked up kinematics equations but I didn't know what to use. Could you possibly show make the steps toward the solution?
     
  5. Mar 17, 2012 #4
    Well, in order to get an exact answer, you would have to include what height the object is dropped from - otherwise there is no way of knowing at what instant your arrow will collide with the bow. What I mean is, if the target is dropped from a very high distance, then if you shoot your arrow straight, you will miss. So, you'll need to aim your arrow up so that it has to cover more distance and slows a great deal, in order to collide with the target exactly

    if the target is dropped from a relatively low distance, you'll probably end up aiming you arrow down so that it races towards the target at such a fast rate that it will impact it before it hits the ground.
     
  6. Mar 17, 2012 #5
    Hey ,

    Have you heard of equations of motion.

    v=u +at
    s=ut+at^2/2
    v^2-u^2=2as

    You have to apply these in two dimensions, x and y.





    Think of the problem this way,

    You are standing 5 m away from where the target is dropped.
    When you shot the arrow, you aimed for the bullseye's initial position.
    However till the arrow covers this 5m, the bullseye falls down a bit due to gravity.
    So you may miss your target.(or is there a catch?)

    You have to shoot in such a way that this fall is compensated.

    Start with this.
    Let initial height of bulls eye be h(from ground)
    And horizontal distance between arrow and bullseye be x.

    If you shoot the arrow at time t=0, let them collide at t=t1.

    In this time the bulls eye moves down(and has motion only in the vertical plane).
    What is the distance moved by it between time t=0 to t=t1.?
    What is its new height from ground?
    (answer in terms of h, g and t1)
    In this time t1, the arrow moves towards the bullseye.
    What are its x and y coordinates at time t1.
    ( what should they be equal to for the objects to collide?)

    (Hint: Equation 2 is to be used)


    (Footnote :
    There is one other phenomenon you need to comsider.
    I am giving you hints right now to help you get closer to the answer.)
     
    Last edited: Mar 17, 2012
  7. Mar 17, 2012 #6

    Hey PotentialE
    How are you ? :-)

    See this is a classic kinematics problem and the results are not similar to what you interpreted when drag is absent.
    Even without knowing the height and distance from where bullseye is dropped and arrow is shot we can tell where exactly the person has to shoot :-)
    The results are amazing and pretty unexpected
     
    Last edited: Mar 17, 2012
  8. Mar 18, 2012 #7

    PeterO

    User Avatar
    Homework Helper

    Very deceptive name for the competition " the feather drop" - since a dropped feather is certainly affected by air resistance - but I expect that in this problem you are supposed to ignore air resistance. Other wise you would have been given some useful information about the size of the air resistance involved.
     
  9. Mar 18, 2012 #8
    Hello, sorry for not being able to solve this own my own with all these hints you have given me. I'm not sure how you use the equation to solve this. Would you mind showing me how to solve it if you don't mind? I haven't taken physics yet so I'm just not sure about what to do.
     
  10. Mar 18, 2012 #9
    Ok.
    I will solve a few steps.

    With reference to my previous post,
    Let velocity of arrow be u at angle m with the horizontal.
    We have find this angle m (which determines where we aim the arrow initially)
    ( assume m is theta.am online via my cell so cant type symbol of theta)

    So for arrow in
    Horizontal direction

    Velocity=ucos(m) towards right
    Acceleration=0 (no force acts)
    Required displacement=x towards right (as for collision the arrow should have sane x coordinate as bullseye )

    Therefore time is t=x/ucos(m)
    (using equation 2)

    In vertical direction

    Initial Velocity=usin(m) upwards
    Acceleration=g downwards (gravity acts)
    Time=x/ucos(m)
    displacement travelled in tine =usin(m) t -gt^2/2

    Plug in t as x/ucos(m)


    .
    Now initially the bullseye was at height h.
    Its velocity initially was 0(it was dropped)
    Its acceleration was g downwards
    After time t
    Its displacement is =-gt^2/2 where negative indicates its downward.

    So its height from ground is h-gt^2/2.


    If the arrow and ball are to collide, this height should be equal to the height of arrow at time t.
    So h-gt^2/2=displcament of arrow.

    What do you see.
    What should be value of m.?
     
  11. Mar 18, 2012 #10
    I don't want to sound stupid or anything, but I still am not completely sure how I solve this problem. I don't want to be a annoying, but could you solve the rest of it so I can see what you did. If not that is completely understandable. I highly appreciate what you have done for me so far.
     
  12. Mar 18, 2012 #11
    Ok.one more hint.

    You will get tan m=h/x.

    Where should the ball be aimed.

    (and relax.I will guide you through the entire solution step by step in case you dint understand.
    Forget the equation for now.you will learn to solve them later
    Just focus on what does tan m=h/x signify.

    Draw a neat diagram and all will be clear
     
  13. Mar 18, 2012 #12
    There is just something about this problem that I just don't understand. Ive drawn pictures and thought about it, but for some reason I am just not able to figure it out.
     
  14. Mar 18, 2012 #13
    Hmm.
    Dont worry.All will be clear.
    Lets begin with this.
    You explain the question in detail in your own words.
    That will help us knkw where you are getting confused.
     
  15. Mar 18, 2012 #14
    I'm not very sure how to set up the equation and use the variables. I am not very good with math and haven't taken any classes that deal with things like sin, cos, and tan. I also am not completely sure how to draw to picture and assign the known values. Once again thanks so much for the help.
     
  16. Mar 18, 2012 #15

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    IF you assume no air resistance (which should not be true for a "feather") then you could just aim your arrow at the target. The force of gravity, and so the downward acceleration, will be the same on both target and arrow so the arrow will go downward at the same rate as the target.
     
  17. Mar 18, 2012 #16
    That makes a lot of sense. I'd still like to know how you solve that mathematically. But still that you so much.
     
  18. Mar 19, 2012 #17
    Hey, solving mathematically can be done only via equations of motion and a brief knlwledge of trigonometry.

    Infact the solution tan (m) = h/x signifies that your initial aim should be exactly where you want to hit the arrow
     
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