A solid cylinder rolls up an inclined plane of angle of inclination `30^(@)`. At the bottom of the inclined plane the centre of mass of the cylinder has a speed of `5ms^(-1)`
How far will the cylinder go up the plane ?

Explanation

Given, `v_c=5ms^(-1),theta=30^(@)`

We know that, `v^2=(2gh)/(1+(K^2)/(R^2))`
Here K`=(R^2)/2` for a solid cylinder
`:.v^2=(2gh)/(1+1/2)=4/3gh" or "h=(3v^2)/(4g)` Taking `g=10ms^(-2),h=(3xx5xx5)/(40)=1.875`
`sin30^(@)=h/l" ":.l=h/(sin30^(@))=1.875xx2`
l=3.75m