The point of contact of sphere has non-zero velocity w.r.t. ground in the forward direction, frictional force will apply in the backward direction. This would decrease both the linear velocity and the angular velocity. frictional force is f=\\mu mg .Linear deceleration is a = f/m = \\mu g Let the time at which pure rolling starts be t .So the linear velocity at this time gives by, v_t = v_o - at = v_o - \\mu gt Angular acceleration is \\alpha = \\dfrac{fR}{2mR^2/5} = 5\\mu g/2R At time t , the angular velocity is given by, \\omega_t = -\\omega_o + \\alpha t = -v_o/2R + 5\\mu gt/2R We have taken angular velocity as negative here to indicate the reverse spin, and angular acceleration positive as it is in the forward spin direction.For pure rolling, v_{t} = \\omega_{t}R\\implies v_o - \\mu gt = -v_o/2 + 5\\mu gt/2\\implies3v_0/2 = 7\\mu gt/2\\implies t = 3v_o/7\\mu g