Find magnitude of two velocities, with one at an angle?

[Brianna I]

Here is the problem:

I am sure I can use pythagorean theorem of quadratic equation to solve but to be honest I'm blanking ridiculously.
I have tried setting it on a coordinate grid and finding the resultant, but I have had no luck! I know I've done this problem before in high school but I'm blanking and have spent a few hours now contemplating this.
Thank you!

 

[Kpgabriel]

You could start by breaking up the wind into its x and y components. Remember your trig.

 

[Brianna I]

You could start by breaking up the wind into its x and y components. Remember your trig.

Hi! I have tried what you said but still do not have the right answer. Here is my work:

Excuse the writing, I'm in a car haha!

 

[Kpgabriel]

Hi! I have tried what you said but still do not have the right answer. Here is my work:
View attachment 106148

Excuse the writing, I'm in a car haha!

Ok, so you found your components for the wind. Now find the components for the glider. look on your graph what angle is the glider traveling?

 

[Kpgabriel]

Ok, so you found your components for the wind. Now find the components for the glider. look on your graph what angle is the glider traveling?

Sorry, it looks like you already have the components for the resultant vector. So you should have no problem finding the magnitude by using Pythagorean theorem.

 

[Brianna I]

Sorry, it looks like you already have the components for the resultant vector. So you should have no problem finding the magnitude by using Pythagorean theorem.

Hi, as you can see I tried that. It did not work :( Do I sum both the x-coordinates? I had 31cos45 - 87 since they are on opposite sides of the coordinate grid.

 

[Brianna I]

Ok, so you found your components for the wind. Now find the components for the glider. look on your graph what angle is the glider traveling?

Ah, would it be at 180^o?

 

[Kpgabriel]

Hi, as you can see I tried that. It did not work :( Do I sum both the x-coordinates? I had 31cos45 - 87 since they are on opposite sides of the coordinate grid.

Right, what did that give you?

 

[Kpgabriel]

Ah, would it be at 180^o?

Yes it would.

 

[Brianna I]

Right, what did that give you?

92.037. I checked, not the right answer (for part a)

 

[Kpgabriel]

92.037. I checked, not the right answer (for part a)

I am getting a different answer. You have r^2 = (-65.08)^2 + (21.92)^2 ?

 

[Brianna I]

I am getting a different answer. You have r^2 = (-65.08)^2 + (21.92)^2 ?

Ah, my mistake. I redid it and got 68.67. I just checked and it is correct! Thank you! I will be able to solve for the angle :)
I must have been too sloppy with the numbers, I realized my calculator was still keeping the negative number even while being squared, messing everything up.
Thanks again!

 

[Kpgabriel]

Ah, my mistake. I redid it and got 68.67. I just checked and it is correct! Thank you! I will be able to solve for the angle :)
I must have been too sloppy with the numbers, I realized my calculator was still keeping the negative number even while being squared, messing everything up.
Thanks again!

You're welcome!

 

[David Lewis]

The problem also asks for the direction of the glider's ground track. (In your sketch, the directional sense of vector R is wrong.) It may help to draw the vectors to scale and connect them head to tail to get a preliminary ballpark resultant.