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Instructor's Algebra Paradox? : How far does a projectile go up a ramp?

  1. Sep 25, 2009 #1
    I'm also working on this other question, which I think my instructor has a paradox in his algebra for lack of what to call it.

    Can anyone suggest a logical way to answer this question?

    1. The problem statement, all variables and given/known data
    http://media.wiley.com/product_data/excerpt/19/04717580/0471758019.pdf
    See problem #43 on page 79

    A ball launched with a velocity of 10.0 m/s at an angle of 50 degrees to the horizontal. The launch point is at the base of a ramp of horizontal length D1= 6.00 m and height d2 = 3.60 m. A plateau is located at the top of the ramp.

    (a) Does the ball land on the ramp or the plateau?

    When it lands, what are the
    (b) magnitude
    (c) angle of it's displacement from the launch point

    The following is all according to my professor

    2. Relevant equations
    Trajectory Equation
    y = tan[tex]\theta[/tex]x - ((g)(x2))/(2vi2)(cos[tex]\theta[/tex])2


    3. The attempt at a solution

    This is what my professor did, but I highly disagree algebraically how this works. The answer is correct, but what he basically does is changes the x variable halfway through the equation:

    y = tan[tex]\theta[/tex]x - ((g)(x2))/(2vi2)(cos^2[tex]\theta[/tex])2

    Notice: He divides both sides by x
    y/x = tan[tex]\theta[/tex] - ((g)(x))/(2vi2)(cos[tex]\theta[/tex])2

    Ramp of slope = 3.60/6.00 = y/x

    Notice: What he is GOING to do is say that the left x is 6.0 (as defined by the question), and then solve for the other x. Why would the OTHER x solve to a different number? They should be the same variable!

    3.6/6.0 = tan[tex]\theta[/tex] - ((9.8)(x))/(2vi2)(cos[tex]\theta[/tex])2

    .6 = tan(50) - 9.8x/(2(10cos50)^2)
    .6 = tan(50) - 9.8x/82.64
    .6 = tan(50) - .1186x
    .1186x = tan(50) - .6
    .1186x = .592
    x = 4.99

    Notice: X now equals a different value than it did before, how does that logically make any sense? He now goes to plug the new x value into the trajectory:

    3.6 = tan[tex]\theta[/tex] - ((9.8)(x))/(2vi2)(cos[tex]\theta[/tex])2

    y = tan(50)(4.99) - (244/72.64)
    y = 5.95 - 2.95
    y = 2.99

    Thus what?

    Thus, the ball lands at the coordinates x = 4.99, y = 2.99

    (b) Displacement vector:

    Square root[(4.99)^2 + (2.99)^2]
    = 5.82 m

    (c) tan-1(2.99/4.99) = 31o

    Basically, look at this and tell me any more logical way to solve this question because as far as I'm concerned, he did an illegal algebra maneuver to fudge the right answer.

    Also, assuming you can solve this a different way, can I basically assume the trajectory formula is a fallacy in physics? I have yet to find a use for it that works besides one question (which happens to be the same as all these other ones).
     
  2. jcsd
  3. Sep 25, 2009 #2
    For y/x he plugged in the slope, a ratio of 3.6/6. That doesn't mean that x is equal to 6 though. A better way to do this with that equationg would be to find out what y equaled in terms of x.
    [tex]tan\theta _{slope} = y/x[/tex]
    You can find the angle of the slope with
    [tex] tan\theta = 3.6/6[/tex]
    Then you can use the quadratic formula or just factor out an x and solve.
     
  4. Sep 25, 2009 #3
    I understand what he did conceptually, but if you look at the formula he went from... that isn't what happened algebraically.

    His "x" did change, and i don't know how that would mean his x didn't equal 6 in the first place.

    That's like saying
    x=6
    y=2

    2y = 5x + 6x^2
    2y = x(5 + 6x)
    2y/x = (5 + 6x)

    2(2)/(6) = 5 + 6x

    4/6 = 5 + 6x
    2/3 = 5 + 6x
    2/3 - 5 = 6x
    (-13/3) = 6x
    (-13/18) = x

    However, if I solve for y and graph that equation, I get x to equal (-5/6) and (0) which is NEITHER of those answers.

    Again, I don't see how algebraically that works out.
     
  5. Sep 25, 2009 #4
    x and y in that equation are variables. You know that the projectile will hit the slope when y/x is equal to the slope which is 3.6/6 aka tan[tex]\theta[/tex]. So therefore you can replace y/x with the slope. That doesn't set x=6 though. x can be any number as long as x/y = 3.6/6
    x could be 60, therefore y would be 36. Or x could be 4.99, therefore y would be 2.99.
     
  6. Sep 25, 2009 #5
    Yes, I understand that portion... but the portion that makes that bogus to look at is the fact that implies that

    Slope = tan[tex]\theta[/tex] - gx/2(vicos [tex]\theta[/tex] )^2

    Which... isn't exactly a formula as far as I know. How do you derive the right side (to make his work logical sense)?
     
  7. Sep 25, 2009 #6
    Basically what I'm saying is...

    y/x = (VoyT + .5gt^2 / VoxT)

    Makes sense... but that doesn't equal that tan(theta) stuff that he used... I don't even get how he got that.

    The above formula makes more sense.
     
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