A sphere pure rolls on a rough inclined plane with initial velocity 2.8 ms-1 . Find the maximum  distance on the inclined plane.

Given, initial velocity of sphere,  u=2.8ms−1Acceleration on the inclined plane,a=gsin⁡θ1+K2R2                 …(i)where, K and R are the radius of gyration and radius of sphere, respectively.Moment of inertia of sphere   I=mK2=25mR2For sphere,   K2R2=25Putting this value in Eq. 0), we geta=gsin⁡30∘1+25    ∵θ=30∘( given ) =5g14=257ms−2   ∵g=10ms−2If s be the maximum distance travelled by the sphere, then at maximum distance, v =0.∴ From third equation of motion,  v2=u2−2as0=u2−2as⇒ s=u22a=(2.8)2×72×25=1.09m≃1.38m