the work done in projecting up the sphere=change in potentiol energy of the system=mgr(1-cos@) from the geometry of the figure

1/2mv^2=mgr(1-cos@)+mglsin@

solving it

v= root 2g(r(1-cos@) +lsin@)

b)

1/2m(2v)^2=mgr(1-cos@)+mglsin@+1/2m(v1)^2

mgr(1-cos@)+mglsin@=1/2mv^2

at the top n+mg=mv^2/r

solving the above equations we get the force acting as

6mg(1-cos@+l/rsin@)

c)

let the particle make an angle@ when it leaves the sphere

mgcos@=mv^2/r

mgr(1-cos@)=1/2mv^2

solving it we get cos inverse 2/3

plz rate me if u find me useful

cheers!!!!!!