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PHYSICS PROBLEM-should be easy for AP students

  1. Oct 5, 2005 #1
    PHYSICS PROBLEM--should be easy for AP students

    1. A football is kicked with an initial speed of 22m/s at some angle above the horizontal. It travels 45 meters horizontally before hitting the ground.
    a. At what angle was the ball initially kicked?
    b. How long does it stay in the air?

    Can u solve it using the 4 essential equations--The equations for constantly acelerated straight-line motion.
     
  2. jcsd
  3. Oct 5, 2005 #2

    Pengwuino

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    What work have you done so far?
     
  4. Oct 5, 2005 #3
    I dont understand how I can get the vertical velocity. I tried using vfinal for vertical to be 0. Then I would get the time and multiply by 2, but I can manage to get to that step.
     
  5. Oct 7, 2005 #4

    Päällikkö

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    vx = constant. Can you figure out what it is more precisely? What can you say about vy and vy0 (vy0 means the initial velocity in vertical direction)?

    x = x0 + vx0t + ½at2
    y = y0 + vy0t + ½at2

    These are two of the equations you meant, right?
    They both can be simplified a bit with the information given.
     
    Last edited: Oct 7, 2005
  6. Oct 7, 2005 #5
    The key is to work this out algebraically without using the values given knowing the fact that [itex]v_x = v \cos \theta[/itex] and [itex]v_{y1} = v \sin \theta[/itex] and using the (simplified) equations that Paallikko stated.
     
    Last edited: Oct 8, 2005
  7. Oct 7, 2005 #6
    I'd focus on this equation: [tex]x=v_{x}\cos{\theta}[/tex]. You can solve for the angle now, and do the rest of the problem.
     
  8. Oct 7, 2005 #7
    Just to note: angle alpha ([tex]\alpha[/tex]) and angle theta ([tex]\theta[/tex]) are supposed to be the same single angle.
     
  9. Oct 7, 2005 #8
    Hey there, sorry to jump in the mix but it seems that all the help, while useful is not heading in the right direction. In orderto solve thie problem ELEGANTLY and efficiently, you gotta mess around with your 4 fundamental kinematics equations. The basic idea here is that whenever you have a problem, especially projectile motion problems, work with what you are given, not your way around the givens. For instance, you were given initial velocity and the RANGE (expressed as either Delta x or R). From these two you need to find first the launch angle and then the time in of flight.
    In order to correctly find the laung angle, you must use the FIFTH equation (yes my friend, there is a fifth one). It's the range equation, which gives you the horizontal range from the SAME level (from ground level to ground level, not ground to air and vice versa). The range equation is:
    R=Vo^2*sin(2Theita)/g
    From this equation, you are given Vo and R plus the fact that g is 9.8. Beautiful, only the angle remains. ONce you solve for the angle,then you can plug the solution into the X=VoCos(Theita)t equation to find your time of flight. The second step can be done any which way you like, but the crucial first step was very specific. Hope this verbose drawn out explanation gave you an insight into this and other motion problems.
     
  10. Oct 7, 2005 #9

    arildno

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    DaMastaofFisix:
    Please learn the difference between fundamental equations and derived equations.
    For what I know, OP was given this assignment in order to DERIVE the range equation!
    In that case, the educative aspect behind the assignment gets quite lost with you blurting out the answer without providing the derivation, doesn't it?
     
  11. Oct 7, 2005 #10
    yea man, just set it up like a right triangle with vertical velocity and horicontal, with the actual velocity as the hypotenus...it should be easy to solve then
     
  12. Oct 8, 2005 #11

    Päällikkö

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    I just don't get it, why on earth would you guys post before the original poster asked for more assistance? In this case, practically all the help after my post has been wrong ([itex]v_x \ne \cos \alpha[/itex], [itex]x \ne v_{x}\cos{\theta}[/itex]). Dwellerofholes made an exception and posted the answer without any explanation (even with the explanation it would've been far too much help).

    Both the four "essential" equations of kinematics and the "range equation" are derived equations. The difference is, that in high school you don't have the tools to derive the kinematics equations, but knowing those you can derive the "range equation".
     
  13. Oct 8, 2005 #12
    I was trying to get the OP going on how to derive the "range equation" without giving away the simplifications of the two equations you posted. Where did I go wrong there?
     
    Last edited: Oct 8, 2005
  14. Oct 8, 2005 #13

    Päällikkö

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    Well, neither of your equations use the initial velocity, meaning that vx and vy are between -1 and 1. This is wrong. You also state that vy is a constant, which it certainly is not.
     
  15. Oct 8, 2005 #14
    Woops, I don't know why I didn't see that. Sorry for any inconvenience!

    I should have used better subscripts - this is what you mean [itex]v_{y1} = v \sin \theta[/itex], no? So v sin theta = 1/2gt.
     
  16. Oct 8, 2005 #15

    Päällikkö

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    To keep the subscripts logical (and because I explained them in my post), I meant: [itex]v_{y0} = v_0 \sin \theta[/itex].

    The hidden part I did not understand. If you place that in the other equation you provided, we'd get that [itex]v_{y1}[/itex] (whatever it means, I'll just assume you mean [itex]v_{y0}[/itex]) does not depend on the initial velocity. Please clarify. What is [itex]v_{y1}[/itex] and what is [itex]v[/itex]?
     
  17. Oct 8, 2005 #16
    I meant [itex]v_{y0} = v_0 \sin \theta[/itex] as you suspected (sorry, again. :blushing:)

    So,

    [tex]
    \begin{align*}
    y &= y_0 + v_{y0}t + \frac{1}{2}at^2 \\
    y - y_0 &= v_{y0}t + \frac{1}{2}at^2 \\
    0 &= (v_0 \sin \theta)t - \frac{1}{2}gt^2 \\
    -(v_0 \sin \theta)t &= -\frac{1}{2}gt^2 \\
    (v_0 \sin \theta) &= \frac{1}{2}gt \\
    \end{align*}
    [/tex]
     
    Last edited: Oct 8, 2005
  18. Oct 8, 2005 #17

    Päällikkö

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    This is a bit irrelevant, but you must neverever divide by zero, and that includes variables. In this case t=0 is an instant when y-y0 = 0.
    The (mathematically) correct way would therefore be:
    [tex]
    \begin{align*}
    y &= y_0 + v_{y0}t + \frac{1}{2}at^2 \\
    y - y_0 &= v_{y0}t + \frac{1}{2}at^2 \\
    0 &= (v_0 \sin \theta)t - \frac{1}{2}gt^2 \\
    0 &= \left( v_0 \sin \theta - \frac{1}{2}gt \right)t\\
    &\Rightarrow t = 0 \vee t = \frac{2 v_0 \sin \theta}{g} \\
    \end{align*}
    [/tex]

    In the OP's problem t = 0 isn't a particularly exciting instant, so we may "ignore" that.
     
    Last edited: Oct 8, 2005
  19. Oct 8, 2005 #18
    Why can't y-y0 equal 0 at the end of the trajectory where t isn't 0 or is that what you were saying in your last sentence?
     
    Last edited: Oct 8, 2005
  20. Oct 8, 2005 #19

    Päällikkö

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    It does (in this problem), as you can see from my solution above. The point was that it equals 0 at t=0, too.
     
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