# A dart is thrown

1. Sep 17, 2016

### Kpgabriel

1. The problem statement, all variables and given/known data
A dart is thrown horizontally toward the bull's-eye, point P on the dart board in the figure, with an initial speed of 8.9 m/s. It hits at point Q on the rim, vertically below P 0.20 s later.
a) What is the distance PQ?
b) how far away from the dart board did the dart thrower stand?
2. Relevant equations
x=xi + .5(vi+vf)t
vf^2= vi^2 + 2a(xf-xi)

3. The attempt at a solution
I am not really sure how to begin this problem. I know the knowns are
vi = 8.9 m/s
t= 0.20 s
vf = 0 m/s
And I believe I need to use one of the kinematic equations.

2. Sep 17, 2016

### hmmm27

Can we start with the basics. What do those Relevant Equations represent, and where should they be used ?

To help a bit further, $v_f$ is not zero. We want velocity when the dart arrives, not when it's embedded in the board.

Last edited: Sep 17, 2016
3. Sep 17, 2016

### Kpgabriel

I got the distance away from the board solved but I do not think that final velocity is needed to find the y-component from P to Q. Instead, I tried to solve for the variables related to y and got.
vi = 0m/s
a = -9.8 m/s^2
t = 0.20s
Δy = ?

4. Sep 17, 2016

### hmmm27

Okay, so what equation should be used, that you can plug those numbers into to get your Δy.

5. Sep 17, 2016

### Kpgabriel

you were right about the velocity. I found it with respect to y using v = vi +at and then substituting into Δy = .5(vi+vf)t which gave me the answer.

6. Sep 17, 2016

### hmmm27

k, as long as you didn't use the same $v_i$ in 'a' as 'b'. You could have gone straight to $d=v_it+at^2/2$

Last edited: Sep 17, 2016