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Projectile motion problem (find the angle) other ways?

  1. May 19, 2013 #1
    1. The problem statement, all variables and given/known data

    A soccer player is taking a penalty kick from a distance of 11 meters from a goal which is 2.44 m high (forgive the possible incorrect use of English, since I'm not a native speaker). Given an initial speed of the ball of 90 km/h (which is 25 m/s2) calculate the initial upward angle of the ball for it to hit the upper bar (i.e., have height = 2.44 after having traveled 11 m horizontally).

    2. Relevant equations

    The trajectory equation for [itex]x_0=y_0=0[/itex] (initial horizontal and vertical positions) is:

    [itex]y = (tan\theta)x - (\frac{g}{2(v_0)^2 cos^2 \theta})x^2[/itex]

    Some known constants in the above equation are:

    [itex]v_0=25 m/s \ \ \ \ \ \ g = 9.81 m/s^2[/itex]

    And obviously [itex]\theta\in(0,\frac{\pi}{2})[/itex]


    3. The attempt at a solution

    The straightforward way is to just substitute the known data in the main equation and solve for [itex]\theta[/itex]:

    [itex]2.44 = 11(tan\theta) - (\frac{9.81}{2(25)^2 cos^2\theta})11^2[/itex]

    But I'm lazy and I find solving for theta to be long and tedious (for me at least) so what I'm asking is:

    Are there are any other ways for finding the angle in the mentioned problem besides the classic method I gave you?

    I feel it's a long shot but I'm in no hurry and it doesn't hurt to ask. Also it doesn't necessarily have to be an easier method. Just post any you can think of.

    Thanks for reading this and hopefully answering it.

    Have a nice day! :smile:
     
  2. jcsd
  3. May 19, 2013 #2

    SammyS

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    Are you too lazy to solve a quadratic equation?

    To get to that point will require using a couple of trig identities and doing some basic algebra.
     
  4. May 19, 2013 #3
    Well, if you have a graphing calculator like a TI84, you might be able to plot the function and look for a zero.
     
  5. May 20, 2013 #4
    I guess someone should've told me that before playing with sines and cosines for half an hour. Anyway, I looked it up on the Internet (and this post is part of my 'research') and I suppose you're referring to this.

    I'll have to admit that it would not have occurred to me to substitute 1/cos^2 with tan^2 + 1 but once you do that the rest is just algebra, as you say.

    Thank you very much for the hint.
     
  6. May 20, 2013 #5
    This could have been solved almost without any trig by considering the vertical and horizontal motions separately.
     
  7. May 20, 2013 #6
    This made me curious. How so?

    Given: [itex]v_y = v_0 sin \theta \ \ \ \ \ v_x = v_0 cos \theta[/itex]

    Vertical motion: [itex]y = y_0 + v_y t + \frac{1}{2}at^2 \ \ \ \ \rightarrow y = v_0 (sin \theta)t - \frac{1}{2}g t^2[/itex]

    Horizontal motion: [itex]x = x_0 + v_x t + \frac{1}{2}at^2 \ \ \rightarrow x = v_0 (cos \theta)t[/itex]

    How would you have solved it differently?
     
  8. May 20, 2013 #7
    Stick with ##v_x## and ##v_y## till the end. When you find them, you will obtain ##\theta##, this being the only time you have to use trigonometry, but it will be of a most basic kind.
     
  9. May 20, 2013 #8
    Shoot the monkey

    Maybe this will also do?
     

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  10. May 20, 2013 #9
    I think powerof's equations are correct.
     
  11. May 20, 2013 #10

    verty

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    I think this problem may be just inherently difficult.

    To get a general formula for ##\tan{θ}##, make the substitution ##k = \frac{v_0^2}{g}##. After much algebra, a not too hideous formula results.
     
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