Question

A dart player throws a dart horizontally at a velocity of 12.4 m/s. The dart hits the board 0.32m below the height at which it is thrown. How far way is the player who threw the dart.

Answer 1

will i get
a medal if i anwser

Answer 2

yes if you can show your work clearly

Answer 3

okay

Answer 4

12.72

Answer 5

am i correct??

Answer 6

I need help, I don't know if your correct

Answer 7

is it an optional question?

Answer 8

yes

Answer 9

give me the options

Answer 10

@CoconutJJ would you like me to help you?

Answer 11

YES!

Answer 12

hey thats hurts

Answer 13

@MinecrodMe you weren't offering a good way to solve it, that's why I butt in that's all...

Answer 14

anyways so what we really need to do is solve for the amount of time it takes the dart to drop that much. Do you know how to do so?

Answer 15

0^2 = vi^2 + 2(-9.8)(0.32)

I solved it like this for initial velocity

Answer 16

giving me vi = 2.5m/s

Answer 17

okay

Answer 18

A hiker started her hike at 0.2 mi below sea level. At the end of her hike, she was 1.4 mi above sea level.

What vertical distance did she hike?

miles

Answer 19

hmmm I think that you got the formula incorrect. What is the formula that you are using?

Answer 20

woops forgot 0.32^2

Answer 21

wait so its 1.416.. m/s

Answer 22

hmmm I am still a bit puzzled at what you did so:
You should have used this formula:
\[x = x_{0}+v_{x}t+1/2(a)t^{2}\] did you do so?

Answer 23

yes...

Answer 24

solving for vi

Answer 25

so what did you put in for each value?

Answer 26

\[y = vt +\frac{ 1 }{ 2 }at ^{2}\]

\[t = \sqrt{\frac{ 2y }{ g }}\]
\[t = \sqrt{\frac{ 2(-0.32m) }{ -9.8m/s ^{2} }}= 0.25s\]
\[x = vt + \frac{ 1 }{ 2 }at ^{2}\]
\[x = vt +\frac{ 1 }{ 2 }(0)t ^{2}\]
x = (12.4m/s)(0.25s) = 3.1m

Answer 27

could you help me with my question??

Answer 28

i will give a medal

Answer 29

@MinecrodMe you need to open your own question, don't post your question on other people's post.

Answer 30

@CoconutJJ so we don't need to solve for vx, since we actually know it

Answer 31

what we need to solve for is t

Answer 32

vi = 12.4m/s

Answer 33

oh I am sorry yes you are correct, we do know what vx is, (I actually wrote the wrong equation, this is the equation we need to use)
\[y=y_{0}+v_{y}t+1/2(a)t^{2}\]

Answer 34

Do you see what I did?

Answer 35

or are you just extremely confused at this point?

Answer 36

yes

Answer 37

ok I think that yes is meant for the first question I posed. Anyways can you tell me the values of these variables:
1) y
2) y0
3) vy
4) a
?

Answer 38

y = 0.32m

Answer 39

y0 = 0m

Answer 40

vy = 12.4m/s

Answer 41

a = 9.8m/s

Answer 42

horizontally at a velocity of 12.4 m/s...vy = 0m/s

Answer 43

ooh there's one thing that wrong. So vy doesn't equal 12.4m/s, vx is what equals 12.4m/s. We are told that it's the horizontal velocity.

Answer 44

@starj9 gonna give u a medal for free :)

Answer 45

oh... I see..

Answer 46

ok good! So now can you solve for t? aka how long it took for the dart to drop that far?

Answer 47

so 0t +1/2(9.8)(t) = 0.32?

Answer 48

You know if yo looked up I already solved this step by step...

Answer 49

you forgot the square on the t

Answer 50

oh yea..

Answer 51

so what is t?

Answer 52

\[t=\sqrt{\frac{ 0.32 }{(0.5)(9.8) }}\]

Answer 53

yes! ok so keep that value. Now think about this, the amount of time it took to drop that far, is the same amount of time it took the dart to go from your hand to the dart board. Does this make sense?

Answer 54

hmm... wait how?

Answer 55

so when a dart leaves your hand it's both going forwards (due to it's initial velocity) and downwards (due to gravity). When it hits the dartboard, it stops, both dropping and going forward at the same time right?

Answer 56

yes..

Answer 57

Because it's embedded into the dartboard. Now do you agree that the darts starts dropping and going forward at the same time?

Answer 58

yes

Answer 59

do you agree that it stops going forward and downward at the same time?

Answer 60

yes

Answer 61

so the time must be the same! So the amount of time it took to drop that far is the amount of time it took to travel that far to the dartboard. Make sense?

Answer 62

now it does..:)

Answer 63

ok good. Now we know the time it took to travel that far. And we know the speed of the dart. So we use the same equation as above(but this time with x-components): \[x = x_{0}+v{x}t+1/2(a)t^{2}\]
now tell me what variable we know and the values we put for them, and what we need to solve for.

Answer 64

oops the equation should ne \[x= x_{0}+v_{x}t+1/2(a)t^{2}\]

Answer 65

vx = 12.4m/s
t = 0.2555s
a = 9.8 ??

Answer 66

ahhh you are right to be a bit puzzled by the a, so the a as you probably know stands for acceleration. Now when we were looking at the dropping we knew that the dart was accelerating downwards due to gravity. But for moving horizontally does gravity affect you? aka does gravity pull/push you forwards or backwards?

Answer 67

ohh i see, so a=0?

Answer 68

yes! There's really nothing that is pulling/pushing us forwards or backwards, and thus the acceleration in the horizontal plane is 0.

Answer 69

This question should be posted in the physics section..

Answer 70

so x = 12.4(0.2555) + 1/2(0)(0.2555^2) ?

Answer 71

@CoconutJJ Make a new question for each of the questions you need help on!

Answer 72

yup that's right!

Answer 73

so how far are we from the dartboard?

Answer 74

3.1682m

Answer 75

yup that looks about right!

Answer 76

THANK YOU @1DEA

Answer 77

no problemo

Answer 78

wow haha it censored me

Answer 79

THANK YOU TOO @StarJ9