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Homework Help: If anyone's awake projectile motion!

  1. Jun 12, 2006 #1
    All right, before I look like a big idiot, I want to preface this with a statement that I have been through Cal III, Abstract Algebra, I'm a senior in college, and I got a 620 on the math portion of the SAT.

    And I CANNOT get this basic problem in projectile motion figured out. Physics has humbled me. Anyone that understands physics gets my highest regard after just getting to the third chapter in this algebra-based physics book.

    Here's my problem:

    We've got a football being thrown at an undetermined angle, theta, at 20m/s. It lands (at the same height, for the sake of ease) 25 meters away.

    I know that we know Xnaught, X, Vnaught, asuby, ynaught and y.

    So I start trying to set the problem up and come up with:


    I know that the Vnaughtx (which I can't type, since it looks dirty o:) ) is basically the velocity (20 m/s) times the cosine of the angle of release. Vnaughty (:blushing: ) is the same, but with sine, right?

    Okay, so I solve down algebraicially to (xay)/(-2v^2) = cos(theta) sin(theta)

    And I've got nowhere to go after that. I got nothin'. I checked out the trig identities, thinking that might be the equivalent of Tangent, but it's not.

    Am I using the wrong equation?

    Thanks in advance for anyone who can help me not fail this class.
  2. jcsd
  3. Jun 12, 2006 #2
    The correct equation for the range of a projectile,
    [tex]R = \frac{u^2sin(2\theta)}{g}[/tex]

    If you look closely you will find that you get the same range for [itex]\theta[/itex] and [itex]90-\theta[/itex].
    Can you do the problem now ?
    If you want to know how the equation is derived, I will be happy to give you a walkthrough. :D
  4. Jun 12, 2006 #3
    Thank goodness you were on. I looked through when I saw that and didn't see that equation anywhere in the book. I think since this is an intro physics book based on algebra, we had specific equations for specific instances.

    Okay, I took it down to arcsin(Rg/v^2)/2 = [itex]\theta[/itex] and got 18.9 degrees. Checked with the back of the book, and it looks like it's good. However, it also gives 71.1, which is 90-[itex]\theta[/itex], like you said. I'm not following why that's the case. Is it more of those crazy circle properties?

    Again, thank you SO much for helping me out with that.
  5. Jun 12, 2006 #4


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    The max range for any given initial velocity occurs at [itex] \Theta [/itex] =45 deg. Therefore there will always be 2 angles which result in the same range, one less then 45 the other greater then 45.
  6. Jun 12, 2006 #5
    I guess that makes sense; I just can't picture it in my head.

    Good thing I'm just a psych major, and most of the physics I'll do in my life will involve an open book, a pencil and a paper! ;)
  7. Jun 12, 2006 #6


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    Also, consider the sine function. Sketch a sine graph ([itex]\sin(2\theta)[/itex] vs. [itex]\theta[/itex]), now draw a horizontal line above the x - axis. Now how many times does this line intersect the curve in the period? The interestions are where your solutions lie. However, you calculator only shows the values at [itex]0+ \theta[/itex], not at [itex]90 - \theta[/itex] or any subsequent values.
    Last edited: Jun 12, 2006
  8. Jun 12, 2006 #7
    Hoot, little correction there, maybe just a typo .
    You should plot [itex]sin2\theta[/itex] Vs [tex]\theta[/itex] and then draw the horizontal line to intersect at two points while [itex]\theta[/itex] lies in the interval [itex](0 , \frac{\pi}{2})[/itex] .

    Oh, and chib , you're most welcome .
  9. Jun 12, 2006 #8


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    Thank's arun, it was a typo (otherwise we would only have one value for theta in the interval). I have duely correct it.
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