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RoganSarine
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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?
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
Trajectory Equation
y = tan\thetax - ((g)(x2))/(2vi2)(cos\theta)2
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\thetax - ((g)(x2))/(2vi2)(cos^2\theta)2
Notice: He divides both sides by x
y/x = tan\theta - ((g)(x))/(2vi2)(cos\theta)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\theta - ((9.8)(x))/(2vi2)(cos\theta)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\theta - ((9.8)(x))/(2vi2)(cos\theta)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).
Can anyone suggest a logical way to answer this question?
Homework Statement
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
Homework Equations
Trajectory Equation
y = tan\thetax - ((g)(x2))/(2vi2)(cos\theta)2
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\thetax - ((g)(x2))/(2vi2)(cos^2\theta)2
Notice: He divides both sides by x
y/x = tan\theta - ((g)(x))/(2vi2)(cos\theta)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\theta - ((9.8)(x))/(2vi2)(cos\theta)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\theta - ((9.8)(x))/(2vi2)(cos\theta)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).
