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2-d kinematics, find theta

  1. Apr 29, 2012 #1
    Usually these problems are really simple for me, but this one is making me lose my mind:

    "An archer wishes to shoot an arrow at a target at eye level a distance of 49.0 m away. If the initial speed imparted to the arrow is 69.3 m/s, what angle should the arrow make with the horizontal as it is being shot?"

    theta = cos-1(v0x/v0)

    v0x=d/t

    I just have no idea how to find t. Help please?
     
    Last edited: Apr 29, 2012
  2. jcsd
  3. Apr 29, 2012 #2

    Curious3141

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    Consider the vertical motion and use [itex]v = u + at[/itex] where u is initial velocity, v is final velocity, a is acceleration and t is the time. Remember that when the arrow reaches the target, it's at eye level again (level horizontally with the point from which it was shot). Therefore, by the symmetry of the parabolic motion, its final vertical velocity is equal in magnitude but opposite in direction to the initial vertical velocity. Using that, solve for t, substitute into what you already have and use a trig identity to simplify before solving for θ.
     
  4. Apr 29, 2012 #3
    [itex]v = -u[/itex] where [itex]u = 69.3 m/s[/itex]. So:

    [itex]69.3 = -69.3 +at[/itex]

    I don't know the resultant acceleration so I assume I am doing something wrong in coming up with this equation from what you have said.
     
  5. Apr 29, 2012 #4

    Curious3141

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    No, 69.3m/s is the actual initial speed in the direction of launch.

    What you need is the vertical component. It involves trigonometry, just like how you found the horizontal component.

    When you throw an object vertically upward at a velocity u, it will rise to a maximum height then descend. When it passes your hand, it will be travelling downward with velocity -u. You can see this is so from [itex]v^2 = u^2 + 2as[/itex]. When the object is at the level of your hand, the displacement s = 0, so [itex]v = \pm u[/itex], where [itex]v = u[/itex] occurs at the start, and [itex]v = -u[/itex] when the object passes by your hand on its downward trajectory.

    This amounts to the same thing when you consider the vertical motion alone. Except that you need to properly resolve the initial velocity into vertical and horizontal components.

    Start by drawing a right angled triangle, with the hypotenuse being represented by the initial velocity at an angle [itex]\theta[/itex] to the horizontal. What are the lengths of the other two sides? They represent the respective vertical and horizontal velocities.
     
    Last edited: Apr 29, 2012
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