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Parabolic Problem

  1. Jan 28, 2009 #1
    1. The problem statement, all variables and given/known data
    Hey all, I'm trying to get caught up in my AP Physics class, (the book we're using is Giancoli Physics 6th Edition), so I am making notes and doing the problems at the end of the chapters. (I'm on Ch. 3 right now). I understand the concepts, but can't seem to do the problems! Here is one I am stuck on;
    #53 "William Tell must split the apple atop his son's head from a distance of 27m. When William aims directly at the apple, the arrow is horizontal. At what angle must he aim it to hit the apple if the arrow travels at a speed of 35 m/s?"

    2. Relevant equations

    3. The attempt at a solution
    Well, I drew out a diagram, and drew X and Y components for the Initial velocity (35m/s), but beyond that, I'm lost. Help appreciated!:biggrin:
  2. jcsd
  3. Jan 28, 2009 #2


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    You will need to break the velocity into its x,y components in terms of θ.

    The time to get there will be the Vx/Distance to apple.

    The time in the air going up and down is the same time, so figure from the Vy how long that would be.
    (Hint: won't it be twice the time to max height where Vy is momentarily 0?)

    Since the time must be equal ... then set the two equations equal.

    Now you have everything but θ
  4. Jan 28, 2009 #3
    Apologies for stepping on your toes lowly, but i'll post anyway seen as how i spent the time to write this out!

    I'd approach this problem as follows:

    1. Consider the vertical component of flight. The speed at which it initially travels vertically (at t=0) will be equal to [tex]\nu=35sin\theta[/tex]. Now we know that vertically, the arrow will travel up and down in a parabolic fashion however, the total vertical distance travelled by the arrow when it hits the apple will be 0 as we are told the arrow is on the same level as the apple.

    Now use the first of our equations of motion for uniform acceleration and algebraically rearrange to find the time of flight of the arrow in terms of [tex]\theta[/tex]:

    [tex]& s & = ut + \tfrac12 at^2 \qquad[/tex]


    s represents the total distance travelled.
    u represents the vertical component of velocity, in this case [tex]u=35sin\theta[/tex].
    t represents the time of flight.
    a represents the acceleration of the arrow (think gravity..)

    2. Consider the horizontal component of flight. The speed at which it initially travels horizontally (at t=0) will be equal to [tex]\nu=35cos\theta[/tex]. Now remember, horizontally there is no acceleration! (we disregard air resistance for simplicity)

    Using the equation as previous, we can simplify to find that:

    [tex]& s & = ut [/tex]

    Now u represents the horizontal component of velocity, in this case [tex]u=35sin\theta[/tex]. We also know s(the horizontal distance travelled given in the question) and t in terms of theta from step 1.

    Put it all together, solve for [tex]\theta[/tex] et voila!
  5. Jan 28, 2009 #4
    Ugh. Sorry, but I don't understand.:frown:

    I pretty much understand all this:
    From then on, I just get more steadily lost.

    Edit: This really freakin annoys me cause we went over this in class months ago, and I still can't seem to remember fundamental equations or solve (relatively) simple problems.:frown:
    Last edited: Jan 28, 2009
  6. Jan 28, 2009 #5


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    The simpler approach is to think it through from the basics.

    You know he must launch at an angle or his son celebrates no more birthdays. That means that the velocity of the arrow to get there will be the Vx component of its velocity (35*cosθ) divided by the distance.

    You can figure that time of flight though because you know Vy = g*t, is the time to get to max height, Vy = 0. Now double that for the time to return to the same height. So that means t = 2*Vy/g.

    Now since the time of the 2 equations, (Vx,Vy), must be equal, the only unknown you have is the angle θ, in the trig functions that define Vx,Vy.
  7. Jan 28, 2009 #6
    See, I didn't know that. Why is this?
  8. Jan 28, 2009 #7


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    The problem tells you that the arrow is at the same level as the apple.

    If the arrow shoots at 35m/s horizontally, then to travel 27 m it will take .77s.

    But it will fall x in that time by

    x = 1/2*g*t2 = 1/2*9.8*.6 = 2.9 m

    OK. So the boy is safe, but the apple is untouched.

    So you know he will only launch it at 35m/s and it will then be at an angle up to account for the time drop. So his Vx will be given by 35Cosθ and the Vy will then be 35Sinθ.

    So go from there.
  9. Jan 29, 2009 #8


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    Newton's second law

    Hi doogerjr! :smile:
    'cos good ol' Newton's second law :approve: says since there is no horizontal component of acceleration, the horizontal component of velocity must be constant …

    so Vx is always V0cosθ :wink:
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