Help? Physics Problem!!

A dart is thrown horizontally with an initial speed of 6 m/s toward point P, the bull's-eye on a dart board. It hits at point Q on the rim, vertically below P, 0.27 s later. Neglect air resistance.

(a) What is the distance PQ?
___________ m

(b) How far away from the dart board is the dart released?
___________ m

a:
Vertical motion: initial velocity = 0; acceleration = 1g down; time = 0.27 seconds.

s = ut + 1/2 a x t ^2

s = 0 x t x 1/2 x -9.8 x 0.27^2

s = 35.7cm = 0.357m


b:

Horizontal motion: Initial velocity = 6; acceleration = 0; time = 0.27 seconds.

s = ut + 1/2 at^2

s = 6 x 0.27 + 1/2 x 0 x 0.27^2

s = 1.62m.

Assuming that point P is vertically even with the point at which the dart is released.

Givens:
Vix = 6m/s
Viy = 0m/s (initial velocity is assumed to be entirely horizontal)
ax = 0 (no air resistance to slow it down)
ay = -9.81m/s^2 (acceleration due to gravity)
t = 0.27s
Vfx = 6m/s
Vfy = ?
dx = ?
dy = ?

a) dy = ?
dy = (Viy)t + (1/2)ay(t^2)
dy = (0m/s)(0.27s) + (1/2)(-9.81m/s^2)(0.27s)^2
dy = 0.358m --> 0.36m (2 significant figures)

b) dx = ?
dx = (Vix)t + (1/2)ax(t^2)
dx = (6m/s)(0.27s) + (1/2)(0m/s)(0.27s)^2
dx = 1.62m --> 1.6m (2 sig figs)