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RoganSarine
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Paradoxical Algebra: Trajectory of a squirt of water to hit a fly
Can ANYONE explain to me why 3 perfectly algebraically sound procedures all give me a different answer, where only one gives me the correct answer? Also, can you disprove the 2 wrong procedures (IE: As I said before, tell me why they don't work)?
http://media.wiley.com/product_data/excerpt/19/04717580/0471758019.pdf
Problem #45
Upon spotting an insect on a twig overhanging water, an archer fish squirts water drops at the insect to knock it into the water. Although the fish sees the insect along a straight-line path at angle phi and distance d, a drop must be launch at a different angle theta if its parabolic path is to intersect the insect.
If phi = 36 degrees, d = .900 m (direct path from frog to fly; hence, x = .9cos36 and y = .9sin36), and the launch speed is 3.56 m/s, what angle theta is required for the drop to be at the top of the parabolic path when it reaches the insect?
y = tan[tex]\theta[/tex]x - ((g)(x2))/(2vi2)(cos[tex]\theta[/tex])2
Quadratic Formula
[tex] v^2 = v_0^2 + 2 a \Delta x[/tex]
y = vertical distance
y = .9sin36
x = horizontal distance
x = .9cos36
g = 9.8
VInitital: 3.56
.9sin36 = tan[tex]\theta[/tex](.9cos36) - ((9.8)(.9cos36)^2)/(2(3.56)^2)(cos^2[tex]\theta[/tex])
.529 = .726tan[tex]\theta[/tex] -((5.20)/(25.3))(1+tan^2[tex]\theta[/tex])
.529 = .726tan[tex]\theta[/tex] - ((.206)(1+tan^2[tex]\theta[/tex]))
.529 = .726tan[tex]\theta[/tex] - (.206 + .206tan^2[tex]\theta[/tex])
.529 = .726tan[tex]\theta[/tex] - .206 - .206tan^2[tex]\theta[/tex])
.206tan^2[tex]\theta[/tex] - .726tan[tex]\theta[/tex] + .735 = 0
Quadratic formula: No Real Answer
Doing the same thing above, but changing up the way I tackle the equation slightly... which is algebraically sound, but provides a different answer (Wrong answer albeit):
Here I even tried number crunching it differently:
y = tan[tex]\theta[/tex]x - ((g)(x2))/(2vi2)(cos[tex]\theta[/tex])2
y/x = tan[tex]\theta[/tex] - ((g)(x))/(2vi2)(cos[tex]\theta[/tex])2
(.9sin36/.9cos36) = tan[tex]\theta[/tex] - g(.9cos36)/(2(3.56)2)*(1(cos[tex]\theta[/tex]))
.4755 = tan[tex]\theta[/tex] - .2815(1 + tan2[tex]\theta[/tex]
-.2815tan2[tex]\theta[/tex] + tan[tex]\theta[/tex] -.2815 - .4755
-.2815tan2[tex]\theta[/tex] + tan[tex]\theta[/tex] -.757
x = 3.24, .308
Why do I get an answer (the wrong answer) doing it THIS way which is essentially the EXACTLY same thing I just did?
How to get the "right answer", but why do I get a different answer than the previous 2 solutions?
[tex] v^2 = v_0^2 + 2 a \Delta x[/tex]
0 = 3.56sin[tex]\theta[/tex]^2 + 2(-9.8)(.9sin36)
0 = 3.56sin[tex]\theta[/tex]^2 - 10.37
10.37 = 3.56sin[tex]\theta[/tex]^2
3.22 = 3.56sin[tex]\theta[/tex]
3.22/3.56 = sin[tex]\theta[/tex]
= 64.8 degrees (The REAL answer)
Once again, back to square one of how these 3 formulas give me separate answers, but are all sound to me in theory.
Can ANYONE explain to me why 3 perfectly algebraically sound procedures all give me a different answer, where only one gives me the correct answer? Also, can you disprove the 2 wrong procedures (IE: As I said before, tell me why they don't work)?
Homework Statement
http://media.wiley.com/product_data/excerpt/19/04717580/0471758019.pdf
Problem #45
Upon spotting an insect on a twig overhanging water, an archer fish squirts water drops at the insect to knock it into the water. Although the fish sees the insect along a straight-line path at angle phi and distance d, a drop must be launch at a different angle theta if its parabolic path is to intersect the insect.
If phi = 36 degrees, d = .900 m (direct path from frog to fly; hence, x = .9cos36 and y = .9sin36), and the launch speed is 3.56 m/s, what angle theta is required for the drop to be at the top of the parabolic path when it reaches the insect?
Homework Equations
y = tan[tex]\theta[/tex]x - ((g)(x2))/(2vi2)(cos[tex]\theta[/tex])2
Quadratic Formula
[tex] v^2 = v_0^2 + 2 a \Delta x[/tex]
The Attempt at a Solution
y = vertical distance
y = .9sin36
x = horizontal distance
x = .9cos36
g = 9.8
VInitital: 3.56
.9sin36 = tan[tex]\theta[/tex](.9cos36) - ((9.8)(.9cos36)^2)/(2(3.56)^2)(cos^2[tex]\theta[/tex])
.529 = .726tan[tex]\theta[/tex] -((5.20)/(25.3))(1+tan^2[tex]\theta[/tex])
.529 = .726tan[tex]\theta[/tex] - ((.206)(1+tan^2[tex]\theta[/tex]))
.529 = .726tan[tex]\theta[/tex] - (.206 + .206tan^2[tex]\theta[/tex])
.529 = .726tan[tex]\theta[/tex] - .206 - .206tan^2[tex]\theta[/tex])
.206tan^2[tex]\theta[/tex] - .726tan[tex]\theta[/tex] + .735 = 0
Quadratic formula: No Real Answer
Doing the same thing above, but changing up the way I tackle the equation slightly... which is algebraically sound, but provides a different answer (Wrong answer albeit):
Here I even tried number crunching it differently:
y = tan[tex]\theta[/tex]x - ((g)(x2))/(2vi2)(cos[tex]\theta[/tex])2
y/x = tan[tex]\theta[/tex] - ((g)(x))/(2vi2)(cos[tex]\theta[/tex])2
(.9sin36/.9cos36) = tan[tex]\theta[/tex] - g(.9cos36)/(2(3.56)2)*(1(cos[tex]\theta[/tex]))
.4755 = tan[tex]\theta[/tex] - .2815(1 + tan2[tex]\theta[/tex]
-.2815tan2[tex]\theta[/tex] + tan[tex]\theta[/tex] -.2815 - .4755
-.2815tan2[tex]\theta[/tex] + tan[tex]\theta[/tex] -.757
x = 3.24, .308
Why do I get an answer (the wrong answer) doing it THIS way which is essentially the EXACTLY same thing I just did?
How to get the "right answer", but why do I get a different answer than the previous 2 solutions?
[tex] v^2 = v_0^2 + 2 a \Delta x[/tex]
0 = 3.56sin[tex]\theta[/tex]^2 + 2(-9.8)(.9sin36)
0 = 3.56sin[tex]\theta[/tex]^2 - 10.37
10.37 = 3.56sin[tex]\theta[/tex]^2
3.22 = 3.56sin[tex]\theta[/tex]
3.22/3.56 = sin[tex]\theta[/tex]
= 64.8 degrees (The REAL answer)
Once again, back to square one of how these 3 formulas give me separate answers, but are all sound to me in theory.
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