We have two cylindrical objects, A (solid) and B (hollow), rolling down a ramp without slipping. We need to determine the speed of object A at the bottom of the ramp given its mass, radius, and the height of the ramp.
For a rolling object, the total mechanical energy is conserved. The potential energy at the top of the ramp is converted into translational and rotational kinetic energy at the bottom.
The potential energy at the top is:
\[
PE = mgh
\]
where \( m = 0.33 \, \text{kg} \), \( g = 9.81 \, \text{m/s}^2 \), and \( h = 0.135 \, \text{m} \).
The kinetic energy at the bottom is the sum of translational and rotational kinetic energy:
\[
KE = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2
\]
For a solid cylinder, the moment of inertia \( I \) is:
\[
I = \frac{1}{2}mr^2
\]
The angular velocity \( \omega \) is related to the linear velocity \( v \) by:
\[
\omega = \frac{v}{r}
\]
Equating the potential energy at the top to the kinetic energy at the bottom:
\[
mgh = \frac{1}{2}mv^2 + \frac{1}{2}\left(\frac{1}{2}mr^2\right)\left(\frac{v}{r}\right)^2
\]
Simplifying:
\[
mgh = \frac{1}{2}mv^2 + \frac{1}{4}mv^2
\]
\[
mgh = \frac{3}{4}mv^2
\]
Canceling \( m \) from both sides and solving for \( v \):
\[
gh = \frac{3}{4}v^2
\]
\[
v^2 = \frac{4gh}{3}
\]
\[
v = \sqrt{\frac{4gh}{3}}
\]
Substituting the known values:
\[
v = \sqrt{\frac{4 \times 9.81 \times 0.135}{3}}
\]
\[
v \approx \sqrt{1.7646}
\]
\[
v \approx 1.3285 \, \text{m/s}
\]
The speed of object A at the bottom of the ramp is \(\boxed{1.3285 \, \text{m/s}}\).