Question

Please help me. Solve for V

Answer 1

The quantities are multiplied.

Answer 2

\[\Large\rm y=(\tan \theta)x\cdot\frac{-g x^2}{2v_o^2\cos^2\theta}\]We want to solve for v0?

Answer 3

Correct. Thanks for making it look so nice!

Answer 4

I got \[\frac{ -gx^2 }{ 2*(y-\tan \Theta x)*(\cos^2\Theta ) }\]

Answer 5

\[\Large\rm \frac{y}{1}=\frac{g x^3(\tan \theta)}{2v^2\cos^2\theta}\]Yah let's ignore the subscript on v for now >.< that's annoying to carry around.
How bout combine the numerators on the right side like this?
A lil earlier to work with.

Then cross multiply maybe? :)

Answer 6

y minus something? Uh oh :O

Answer 7

I think I am wrong with mine. I am following you.

Answer 8

Cross multiplying gives,\[\Large\rm v^2(2y \cos^2\theta)=-g x^3(\tan \theta)\](I dropped the negative sign in the previous step, my bad)

Answer 9

Ok. Got ya.

Answer 10

So now just divide?

Answer 11

First solve for v^2, divide by the stuff, yes.

Answer 12

You'll have a tangent over cosine squared on the right side.
You might want to simplify that down a little bit.
Maybe not though...

Answer 13

After it is simplified all I have to do is plug in stuff.

Answer 14

\[\Large\rm v^2=\frac{-g x^3(\tan \theta)}{(2y \cos^2\theta)}\]Like you could convert your tangent to sine and cosine, but whatever.\[\Large\rm v=\pm\sqrt{\frac{-g x^3(\tan \theta)}{2y \cos^2\theta}}\]

Answer 15

Gonna plug some stuff in? Ah ok.

Answer 16

Yes. We did a lab with horizontal and vertical motion. Which leads to this: the + and - are not both needed, correct?

Answer 17

Thanks!

Answer 18

I guess you'd have to look at the context of the problem to decide which sign makes more sense.

Negative velocity just means something is moving backwards, right?
So look at your problem and see if that makes sense or not :)
Yah you probably don't need it.

Answer 19

All of my distances are + :)

Answer 20

Humm. I think there is an error in this somewhere. My answers are not coming out like they should. What if cos and tan were in the bottom?