# Solve A Question in 2D Motion A dart player throws a dart horizontally at a spee

1. Oct 24, 2009

### candyvera

1. The problem statement, all variables and given/known data
A dart player throws a dart horizontally at a speed of 11.8 m/s. The dart hits the board 0.31 m below the height from which it was thrown. How far away is the player from the board?

x y
vi= 11.8 vi= 0
a=0 a= 9.81
x= unknown y= unknown
t= unknown t= unknown

2. Relevant equations

V = Vo + at
X = Vot + .5at2
v2 = vo2 + 2a(X - Xo)

3. The attempt at a solution
i started of with the y part
v0t + .5at2
y = 0+.5(9.81)(t2) couldn't solve for the distance since i didn't have the amount of time
and i don't know which equation to use after that because all of them involve velocity and i only have the intial velocity of zero.

2. Oct 24, 2009

### cepheid

Staff Emeritus
Hello candyvera, welcome to PF!

Thankfully, you already know what this distance is. It is given in the problem. You are using this equation to answer the question, "if an object falls 0.31 m while accelerating under gravity, how long does that take?"

3. Oct 24, 2009

### candyvera

You are using this equation to answer the question, "if an object falls 0.31 m while accelerating under gravity, how long does that take?"[/QUOTE]

wait i'm confused, the .31 m is my distance?
i thought the question says that the distance is .31m below the height at which it was thrown?

4. Oct 24, 2009

### cepheid

Staff Emeritus
Exactly. 0.31 m is the distance travelled in the y direction (i.e. the vertical direction). The dart fell 0.31 m during its flight.

5. Oct 24, 2009

### candyvera

okay thnx :] are my calculations correct?
.31 = (0)t + .5(9.81)t2
t= .251

then you multiply the time twice for the x direction

x= (11.8)(.251*2) + .5(0)(.251*2)2
x= (11.8)(.502)
x= 5.93