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Motion along a curved path -- introductory

  1. Oct 28, 2014 #1
    1. "William Tell is said to have shot an apple off his son's head with an arrow. If the arrow was shot with an initial speed of 55m/s and the boy was 15m away, at what launch angle did Bill aim the Arrow? (Assume that the arrow and the apple are initially at the same height above the ground.)"


    2. Relevant equations


    3. The attempt at a solution
    Change in X is 55m/s*cos(theta)*t
    Change in Y is 55m/s*sin(theta)*t - (4.9*t^2)
    Need to solve for either time or theta in order to get the other from what I can see? Is there another way? Or am I wrong all together?

    Didn't mention, I don't need the answer; I have it. I just need a reliable way to solve it. A very in depth-step-by-step with explanations would be very beneficial to me, thank you. And my teacher is also not looking for an answer, so don't fret about "cheating," the assignment was given in order to further understanding of the topic and to be able to solve more similar problems in the future.
     
    Last edited: Oct 28, 2014
  2. jcsd
  3. Oct 28, 2014 #2
    .
     
    Last edited: Oct 28, 2014
  4. Oct 28, 2014 #3

    SteamKing

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    It helps if you write regular equations to describe the motion of the arrow.

    If the arrow is shot at an initial angle, its altitude will increase until gravity arrests its upward motion. Once the upward motion of the arrow ceases, it's going to start to fall back to earth. Simultaneously, the arrow is traveling at constant horizontal velocity towards the target.

    By writing equations which describe this motion, you should be able to solve for both the time it takes the arrow to reach the target and the angle at which it is shot.
     
  5. Oct 28, 2014 #4
    That honestly doesn't help me at all, could you please show some step by step solving? If you need to use a problem with similar goals and different numbers that's fine, but at the moment I'm no better off
     
  6. Oct 28, 2014 #5

    SteamKing

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  7. Oct 28, 2014 #6
    I know those equations, but you need theta to solve for time and vise-versa,
    my question is how to use that information to be able to solve for theta or time.
     
  8. Oct 28, 2014 #7
    Your two equations are:

    ##\Delta X = (55 m/s) \times \cos(\theta)t##
    ##\Delta Y = (55 m/s) \times \sin(\theta)t - (4.9t^{2}) ##

    This is a system of two equations with two variables, you solve through substitution or any other algebraic way of solving systems of equations. One such way is solving for t in the first equation:

    ## t = \frac{\Delta X}{(55 m/s)\cos(\theta)} ##

    And plugging it into the second equation:

    ##\Delta Y = (55 m/s) \sin(\theta)\frac{\Delta X}{(55 m/s)\cos(\theta)} - (4.9(\frac{\Delta X}{(55 m/s)\cos(\theta)})^{2}) ##

    And work from there, solving for ##\theta##, but this way is rather messy. I would suggest doing it using the other substitution, (what is ##\Delta Y##? and how can you solve for t (think factoring or quadratic formula) in terms of ##\sin(\theta)##?). Plugging it in to the first equation, you should get some equation with either the cotangent or tangent of the angle equaling some number and you can take the inverse cotangent or inverse tangent of that number.

    Hope this helps.
     
  9. Oct 28, 2014 #8
    • Sarcastic comments about other members are unacceptable at Physics Forums
    Thank you , that actually did help, but don't say "plugging it in", it's substitution :p anyways thank you; they oughta make you a helper person, not King ambiguous Steam
     
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