A solid sphere is projected up along an inclined plank of inclination `theta=30^@` with a speed `v=2ms^(-1)`. If it rolls without slipping find the maximum distance travelled by it (`g=10 ms^(-2))`

Explanation

Let the solid sphere of mass `m` and radius `r` tranverse a distance `l` along the inclined plane. Since it rolls without sliding its initial `KE` is given as

`KE_(i)=1/2mv^(2)(1+(k^(2))/(r^(2)))`, where `k=sqrt(2/5)r`
`=1/2mv^(2)(1+2/5)k=(7/10)mv^(2)`
As it comes to rest attaining a height `h=lsintheta`, its final `KE =0`
`/_\KE=-7/10mv^(2)`..........i
Change in gravitational `PE=mgh`...........ii
Conservation of energy yields `(/_\PE)+(/_\KE)=0`
`implies mgh-(7/10)mv^(2)=0`
`implies h=(7v^(2))/(10g)(7xx(2)^(2))/(10xx10)=0.28m`
`:. =h cosec theta-0.28 cosec 30^@=0.56m`