desperate need of help! (frictionless slope problem)?

Question

a 16 kg block slides down a frictionless slope which is at an angle of 27 degrees. Starting from rest, the time to slide down is t=1.77 seconds. (acceleration of gravity is 9.8)

1] what total distance did the block slide?
2] what is the total vertical height through which the block descended?

i worked it out and seemed to get a good answer but it was wrong, ahhhhh!

Answer 1

a = g*sin(27) = 4.45 m/s^2
t = 1.77s
v0 = 0

x = x0 + v0t + 0.5at^2

x = 0+0+ 0.5*4.45*1.77^2

x = 6.97 meters - total distance

vertical height y = x*sin27 = 3.16m

Answer 2

The slope being frictionless, the mass of the block is immaterial.
1)
a = gsin27°
s = s0 + v0t + (1/2)at^2
s = 0 + 0 + (1/2)(1.77^2)(9.8)sin27°
s = 6.9693 m
2)
y = ssin27°
y = 6.9693sin27°
y = 3.1640 m

Answer 3

You use the formula: s = u t + 0.5 a t x t to find s for part 1
u =0, a = 9.8 x sin (27degree) -- only this portion of gravity force is pushing the block down, t = 1.77

{ans = 6.969}

Part 2 is simply part 1 x sin(27degree) =3.164m