Uniform solid disk rolls without slipping down a 19.0° inclined plane. what is the acceleration of the disk's center of mass?

Below is the solution:

Let us say that the disk goes through a vertical elevation change of one meter. 

The change in potential energy will equal the change in kinetic energy 

PE = KEt + KEr 
mgh = ½mv² + ½Iω² 

for a uniform disk, the moment of inertia is 
I = ½mr² 
and 
ω = v/r 

mgh = ½mv² + ½(½mr²)(v/r)² 
mgh = ½mv² + ¼mv² 
gh = ¾v² 

v² = 4gh/3 

v² = u² + 2as 

if we assume initial velocity is zero 

v² = 2as 
a = v² / 2s 

s(sinθ) = h 
s = h/sinθ 

a = 4gh/3 / 2(h/sinθ) 
a = ⅔gsinθ 

a = ⅔(9.8)sin25 
a = 2.8 m/s²