A sphere of radius 20 cm and mass 45 g is placed atop a ramp of height 0.75 m and inclination angle 30 deg. If the ramp were frictionless the sphere would slide down the ramp in a time t . With friction, the sphere rolls without slipping down the ramp and reaches the bottom at time t' . What is (t')/t ?

First use conservation of energy to find v at the bottom of the ramp when there is no friction.
E_i=E_f
mgh=1/2mv^2
v=sqrt(2gh)
Now plug into a kinematic equations to find t after the sphere has rolled down the ramp a distance we will call Delta x .
know v^2=2a Delta x
and
v=t*a
Then v*v=2a Delta x
v=(2a Delta x)/v
t*a=(2a Delta x)/v
t=(2Delta x)/v
t=(2 Delta x)/sqrt(2gh)
Now repeat the process with friction to find t' .
E_i=E_f
mgh=1/2m(v')^2+1/2I(omega)^2
I=2/5mR^2 for a solid sphere.
mgh=1/2m(v')^2+1/2(2/5mR^2)((v')/R)^2
gh=1/2(v')^2+1/5(v')^2
gh=(1/2+1/5)(v')^2
sqrt((10gh)/7)=v'
t'=(2Delta x)/(v')=2 Delta x*sqrt(7/(10gh))
Finally, we have
(t')/t=(2 Delta x sqrt(7/(10gh)))/((2 Delta x)/sqrt(2gh))
(t')/t=sqrt(7/(10gh))*sqrt(2gh)
(t')/t=sqrt(7/5)
http://hyperphysics.phy-astr.gsu.edu/hbase/isph.html