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Projectile motion of an arrow

  1. Jun 26, 2010 #1
    1. The problem statement, all variables and given/known data

    You are watching an archery tournament when you start wondering how fast an arrow is shot from the bow. Remembering your physics, you ask one of the archers to shoot an arrow parallel to the ground. You find the arrow stuck in the ground 55.0 m away, making a 3.00 degree angle with the ground.


    2. Relevant equations

    I used basic 2d kinematic equations along with a few adapted for projectile motion.

    3. The attempt at a solution

    The height of the archer/height of his arm, was not given so I said tan(3) = H/55, and solved for H, (y0). This gave about 2.9, which would mean that either the guy is on a platform or is a gigantic monster. I figured it was wrong but I went on a little, using the formula y = H - gx2/ 2 V0x

    Sorry about barely using latex
     
  2. jcsd
  3. Jun 26, 2010 #2
    The reason it gave you a large height is because the arrow actually turns downward.
    Remember from the problem that the archer shot it perfectly horizontally.

    Something that you have to assume in order to solve this problem which they didn't tell you, is that the direction of the arrow must be tangent to the direction of its velocity.
    Therefore, the arrow sticking out of the ground basically tells you the final velocity direction (but not magnitude), after you realize this you can just treat the arrow as a point particle with a final velocity making a 3 degree angle with the horizontal.
     
  4. Jun 26, 2010 #3
    I noticed this before, however I am still not quite sure how to use that fact to find the initial velocity. Or well, I tried using vfy = viy - g[tex]\Delta[/tex]t.
    I figure the heat must be getting to me.
     
  5. Jun 26, 2010 #4
    Given:
    [itex]s[/itex] - range of the arrow
    [itex]\alpha[/itex] - angle of the arrow with the horizontal (measured "downwards")
    You should use the following equations
    ------------------------------------------------------------------------------------
    Required:
    [itex]v[/itex] - initial velocity of the arrow in the horizontal direction

    Suggestion for solution:
    The relevant velocity time and position time equations are:

    [tex]
    v_{x} = v
    [/tex]

    [tex]
    v_{y} = -g \, t
    [/tex]

    [tex]
    x = v \, t
    [/tex]

    [tex]
    y = h - \frac{g \, t^{2}}{2}
    [/tex]

    where [itex]h[/itex] is the height from which the arrow was released and is unknown, so it must be an intermediate variable that needs to be eliminated from the given conditions.

    What condition is mathematically expressed with the statement "the arrow hits the ground at time [itex]t_{0}[/itex]"?

    How can you find the range of the arrow at that time?

    How can you find the angle it builds below the horizon?

    From this, you will have 3 equations with 3 unknowns ([itex]t_{0}[/itex], [itex]h[/itex] and [itex]v[/itex]), out of which 2 are intermediate ([itex]t_{0}[/itex], [itex]h[/itex]), so it would be a matter of algebra.
     
  6. Jun 27, 2010 #5
    Vx is constant. Now use Vy/Vx = tan 3. Now u know the time taken will be 55/Vx.
    Also Vy=0.5g (55/Vx)^2. Thus u got the two equation relating Vy and Vx. Hence u can find out the speed of the arrow which is Vx.
     
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