Problem with Bernoulli's equation

[Loopas]

(1)

Hey everyone,

I'm only having trouble with part (b) of this problem. I attached a picture of the problem and its diagram.

(2)

x=vt
the x(t) kinematics equations
Bernoulli's equation

(3)

I set the equations for distance as equal:

2√(y1(y2-y1))=2√(y1'(y2-y1'))

But I don't think this can be solved for y1'. I also have a feeling that I may have to use the quadratic equation but I'm not sure exactly how.

Thanks for the help.

 

[462chevelle]

mathematically you should just be able to distribute the y1 through the parenthesis, then get rid of the radical(provided that is what that symbol is), then solve the quadratic problem. but I cannot tell you if your physics formula is correct. I've not worked with bernoullis other than conceptual problems.

 

[gneill]

(1)

Hey everyone,

I'm only having trouble with part (b) of this problem. I attached a picture of the problem and its diagram.

(2)

x=vt
the x(t) kinematics equations
Bernoulli's equation

(3)

I set the equations for distance as equal:

2√(y1(y2-y1))=2√(y1'(y2-y1'))

But I don't think this can be solved for y1'. I also have a feeling that I may have to use the quadratic equation but I'm not sure exactly how.

Thanks for the help.

The leading 2's cancel on each side. Then square each side. Should make life simpler for you.

 

[Loopas]

That simplifies to:

y1y2-y1^2=y1'y2-y1'^2

Now here is where I'm stuck. I'm not sure how to use this with the quadratic equation or solve for y1'.

 

[gneill]

Well, everything but y1' are constants. Move everything to one side and sort it into the standard form of a quadratic.

 

[Loopas]

Found the answer, thanks for the help!