Question

Help with application of vectors

Answer 1

An arrow is shot into the air so that its horizontal velocity is 30.0 ft/sec and it vertical velocity is 10.0 ft/sec respectively. Find the velocity of the arrow. Round your answer to the nearest ft/sec.


32 feet per second

20 feet per second

40 feet per second

28 feet per second

Answer 2

well I think we need to find the magnitude and the direction

Answer 3

Welp that is easy

Answer 4

sqrt(30^2)+(10)^2=

Answer 5

31.6 ||v||

Answer 6

or ||v||= 10sqrt10

Answer 7

\[v_x \text{ represents } \text{ horizontal velocity } \\ v_y \text{ represents vertical velocity} \\ |v|=\sqrt{x_x^2+x_y^2} \\ \tan(\theta_v)=\frac{v_y}{x_x} \\ \text{ where } |v| \text{ is maginaute of vector } \\ \text{ and } \theta_v \text{ is direction of the object }\]

Answer 8

\[\theta=\arctan(\frac{ 30 }{ 10 }) \]

Answer 9

\[\theta=72 degrees\]

Answer 10

wait no I screwed up

Answer 11

Theta=18 degrees

Answer 12

yeah I have theta is approximate 18 deg

Answer 13

Alright yeh I got messed up with directions.

Answer 14

at least i caught it

Answer 15

so ||10sqrt10||cos18

Answer 16

Nevermind that wont give us the velocity it would just give us the horizontal point

Answer 17

ok and based on my resources we can either calculate
\[\frac{v_x}{\cos(\theta) } \text{ or } \frac{v_y}{\sin(\theta)} \]
(I lied we may not need the mag)

Answer 18

either will give you same answer

Answer 19

http://www.asu.edu/courses/kin335/documents/Trigonometry.pdf
used 5 as an example

Answer 20

so 30/cos18

Answer 21

yeah

Answer 22

32 feet per second?

Answer 23

that is what I think yes
(and I will tell you I based my formula off of that website there for number 5)
like it is kinda the same thing except backwards
it asked for the vertical and horizontal given the angle and the velocity

Answer 24

Yeah I didnt know really where to go with this question. I just knew it gave vertical and horizontal aand we could do somethign with ||v||cos(or sin)Theta

Answer 25

so they had
\[v_y=\text{ velocity } \cdot \sin(\theta) \]
and
\[v_x=\text{ velocity } \cdot \cos(\theta)\]
based on there answers

Answer 26

their*

Answer 27

Lel that is something im always gonna suck at. identifying equations

Answer 28

hey omg

Answer 29

you know I think all we had to do was use pythagorean theorem

Answer 30

we would still get 32

Answer 31

lol