Question

Help!

A superhero flies 160 m from the top of a
tall building at an angle of 20 ◦ below the
horizontal.
What is the horizontal component of the
superhero’s displacement? Draw the vectors
to scale on a graph to determine the answer.
Answer in units of m. Your answer must
be within ± 5.0%

What is the vertical component of the superhero’s
displacement?
Answer in units of m. Your answer must
be within ± 5.0%

I've gotten the first part, it's 150.612 but I can't manage to get the first part, whether it's Pythagoras or sin/cos. 54.anything is not a right answer.

Answer 1

@Directrix the solution looks correct, am i missing anything?

Answer 2

Look at the attachment and let me know if it makes sense. @sooobored

Answer 3

Y is 150.612
the problem is that it says 54.723 is wrong

Answer 4

Vertical Component

https://sehsphysics.wikispaces.com/file/view/summer+quest+2+KEY.pdf

Answer 5

I've gotten 54.723 so many times I want to screaM but it says that it's not correct!

Answer 6

We start out with this correct?

Answer 7

no we only start with the hypotenuse and the angle °

Answer 8

yeah that looks right :)

Answer 9

We can then find x by using cos.

\[\cos{(20)} = \frac{x}{160}\]

Solving for x, we find x = 150.35

We can then find y using the pythagorean theorem.

\[a^2 + b^2 = c^2\]

Answer 10

just the things ive tried that were all wrong

Answer 11

@Austin.L

Should the equation be this: cos (25) = x/160 ?

Answer 12

the angle° is 20

Answer 13

Or rather,
\[x^2 + y^2 = h^2\]

We know, h = 160 & x = 150.35

\[(150.35)^2 + y^2 = (160)^2\]

Solving for y we find y = 54.7255

Answer 14

Okay, the attachments will give an example of how to work the problem but the angle of the attachment problem is not the same as that of this problem. Sorry to mislead.

Answer 15

it's no problem! :)

Answer 16

Or we can go about it using trig.

\[\sin{(20)} = \frac{x}{160}\]

Which when you solve for x is 54.723

I'm not sure why you'd have anything else.

Answer 17

that's what I'm saying! but it always counts it as wrong, and im on my last try here so I don't knwo

Answer 18

I've solved it with trig formulas, and cos and sin... and I get approximately the same answer. I'm not sure why it says its wrong man.

Answer 19

ill try it again, maybe glitching! thanks for your help!! :)

Answer 20

No problem buddy!

Answer 21

Yeah, I feel there might be something missing... I'd rather have you post a screeny of the question next time.

Answer 22

wrong :(

Answer 23

what is like the format

Answer 24

of ur answer input, like does the website check it to some sigfig?

Answer 25

no sigfig checks

Answer 26

That looks about right....so you got 150.351m for horizontal
and -54.723m for vertical...

Answer 27

Have you tried inserting a negative in?
so instead of 54.7
Make it -54.7

Answer 28

Wait... waiiiiiiiit.... It would be a negative!!!

Answer 29

DARN IT!! I put in my last answer so it won't accept any more!! :S

Answer 30

Yeah, happens in 2Dimensional problem.s

Answer 31

oh well, rip 10 points.

Answer 32

Here we are.

Answer 33

I'm so sorry, I forgot that it was 20 degrees down which means the displacement would be down meaning negative! D':

Answer 34

why would it be negative over normal?

Answer 35

Actually it's negative cuz it's below the horizontal, that's it.

Answer 36

If it was 20 degrees above the horizontal, then the vertical component would be positive.

Answer 37

darN!