$A$ uniform spherical object of mass $M$ and radius $R$ has a moment of inertia $I$. It rolls without slipping down an inclined plane of angle $\theta$. What is its acceleration?

$A$ solid spherical ball rolls on a horizontal surface at $10 \ m \ s^{-1}$ and continues to roll up on an inclined surface as shown in the figure. If the mass of the ball is $11 \ kg$ and frictional losses are negligible, the value of $h$, where the ball stops and starts rolling down the inclination is $($Assume $g = 10 \ m \ s^{-2} )$ (in $m$)

$A$ hollow spherical ball of uniform density rolls up a curved surface with an initial velocity $3\, m/s$ (as shown in figure). The maximum height with respect to the initial position covered by it will be $...........cm$.

$A$ ring and a disc are initially at rest, side by side, at the top of an inclined plane which makes an angle $60^{\circ}$ with the horizontal. They start to roll without slipping at the same instant of time along the shortest path. If the time difference between their reaching the ground is $(2-\sqrt{3}) / \sqrt{10} \,s$, then the height of the top of the inclined plane, in metres, is. . . . . . Take $g=10 \,m \,s^{-2}$.

$A$ solid sphere of mass $2 \,kg$ rolls on a smooth horizontal surface at $10 \,m/s$. It then rolls up a smooth inclined plane of inclination $30^{\circ}$ with the horizontal. The height attained by the sphere before it stops is [take $g=10 \,m/s^2$].

$A$ tennis ball (treated as a hollow spherical shell) starting from $O$ rolls down a hill. At point $A$,the ball becomes airborne,leaving at an angle of $30^\circ$ with the horizontal. The ball strikes the ground at $B$. What is the value of the distance $AB$ (in $m$)? (Moment of inertia of a spherical shell of mass $m$ and radius $R$ about its diameter is $I = \frac{2}{3}mR^2$).