Dreadful Rotation! 2

Initially the ball has no spin, and hence it's point of contact with the supporting surface moves forward, and friction acts in the direction opposite to that of motion. The linear acceleration is $0.1\times 10=1$ m/s $^2$ and the angular acceleration is $\dfrac{5\times 0.1\times 10}{2\times 1}=2.5$ rad/s $^2$ . The time $t$ taken to cover the distance of $21560$ m. is given by $21560=343t-\dfrac{1}{2}t^2\implies t^2-686t+43120=0\implies t=70$ sec. After travelling $21560$ m. , the translational velocity of the C.M. of the ball is $273$ m/s. The angular velocity of the ball is $175$ rad/s. The vertical impulse received by the ball on landing on the bottom surface is $\hat {J_v}=\sqrt {2\times 10\times 245}=70$ kg-m/s. The horizontal impulse is given by $\hat {J_h}=273-v_1=\dfrac{2}{5}\omega_1-70$ kg-m/s. , where $v_1$ and $\omega_1$ are the final linear and angular velocities of the ball respectively. With $v_1=\omega_1\times r=\omega_1$ , we get $3.5\hat {J_h}=98\implies \hat {J_h}=28$ . For pure rolling to prevail, we must have $\hat {J_h}\leq \mu\hat {J_v}\implies 28\leq 70\mu\implies \mu\geq \dfrac{28}{70}=\boxed {0.4}$