Dart thrown horizontally hits point below target in 0.20 s

[Kpgabriel]

Homework Statement

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?

Homework Equations

x=xi + .5(vi+vf)t
vf^2= vi^2 + 2a(xf-xi)

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.

 

[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.

 

[Kpgabriel]

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.

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 = ?

 

[hmmm27]

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

 

[Kpgabriel]

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

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.

 

[hmmm27]

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