$A$ solid spherical ball is rolled up an inclined plane of angle of inclination $30^{\circ}$ with an initial speed of $4 \ m/s$ at the bottom of the inclination. How far will the ball go up the plane (in $cm$)? (Use $g=10 \ m/s^2$)

What is the velocity of a sphere starting from rest and rolling without slipping down an inclined plane of vertical height $h$?

Consider a situation in which a ring,a solid cylinder,and a solid sphere roll down on the same inclined plane without slipping. Assume that they start rolling from rest and have identical diameters. The correct statement for this situation is:

The acceleration of a body rolling down on an inclined plane does not depend upon

$A$ sphere,a disc,a ring,and a hollow sphere of the same radius are rolled down from the same height on an inclined plane simultaneously. The order in which these objects reach the bottom is:

$A$ solid sphere of mass $1 kg$ and radius $1 m$ rolls without slipping on a fixed inclined plane with an angle of inclination $\theta = 30^{\circ}$ from the horizontal. Two forces of magnitude $1 N$ each,parallel to the incline,act on the sphere,both at a distance $r = 0.5 m$ from the center of the sphere,as shown in the figure. The acceleration of the sphere down the plane is . . . $m s^{-2}$. (Take $g = 10 m s^{-2}$.)