Solve for v: "How to Solve for v with Theta, g, and x

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The discussion focuses on solving for the variable v in the equation theta = tan^-1((v^2 - (v^4 - g(gx^2))^1/2) / gx), with specific values of x = 30, theta = 45 degrees, and g = 9.81 m/sec². Participants suggest rewriting the equation to isolate v² and squaring both sides to eliminate the square root, leading to a quadratic equation in terms of u = v². They emphasize the importance of checking solutions due to potential extraneous roots introduced by squaring the equation.

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Homework Statement


theta = tan^-1 {{v^2 - [v^4-g(gx^2)]}^1/2}/gx
I know that x is 30, theta is 45, g is 9.81 m/sec^2.
I'm trying to solve for v.

Homework Equations


The Attempt at a Solution


I've gotten all the way down to
v^2 = (v^4-86612.49)^1/2 + 4.165
What do I do from here?
 
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marthkiki said:

The Attempt at a Solution


I've gotten all the way down to
v^2 = (v^4-86612.49)^1/2 + 4.165
What do I do from here?

you'll need to re-write it as


v2 - 4.165= (v4-86612.49)1/2

then square both sides and solve. You may need to use a numerical method. You'll need to test your solutions as squaring both sides might give you false answers.
 
Once you square it you will be left with only 1 V term. It may be easier to replace your constants with variables. Then just substitute them in once you find V.
 
How "only 1 V term"? There will be v^4 and v^2 terms but then you can let u= v^2 and will have a quadratic to solve for u.
 
rock.freak667 said:
you'll need to re-write it asv2 - 4.165= (v4-86612.49)1/2

then square both sides and solve. You may need to use a numerical method. You'll need to test your solutions as squaring both sides might give you false answers.

By squaring both sides, I get v^2 - 17.35 = v^4 - 86612.49.
That gets me -17.35 = -86612.49.
Is this what you mean by a false answer?
 
HallsofIvy said:
How "only 1 V term"? There will be v^4 and v^2 terms but then you can let u= v^2 and will have a quadratic to solve for u.

If i do that, I end up with u = v2 + 416.2
then the discriminant will end up negative leaving me with an imaginary velocity. Did I do this correctly?
 
If i do that, I end up with x = 86612.49 and y = 4.165. I then end up with v^4 - y^2 = v^4 - x, which again gets me to -y^2=x.
 
v^2 -4.165 = (v^4-86612.49)^1/2
v^2 - x = (v^4-y)^{1/2}
v^4 - 2v^2x + x^2=v^4-y
-2v^2x + x^2=-y
v=\sqrt{\frac{x^2+y}{2x}}
there u go
 
marthkiki said:
By squaring both sides, I get v^2 - 17.35 = v^4 - 86612.49.
The left-hand side is not squared properly. The left hand side should be (v^2 - 17.35)^2 = v^4 - 2(17.35)v^2 + (17.35)^2. The v4's will then cancel and you will be left with a quadratic in v. You will need to check the two answers you get, as squaring both sides of an equation introduces new solutions that may not solve the original equation. Ie., x = -1 and x2 = 1, which introduces the solution x = 1.
 
  • #10
Thank you so much!
I get it now. I totally squared the left wrong.
 

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