Solution:

Since \(3/4^{\text{th}}\) of kinetic energy is lost, the remaining kinetic energy is \(1/4^{\text{th}}\), meaning the final velocity \(v = v_0/2\). Using \(v = v_0 - at_0\) where \(a = \mu g\), we get \(v_0/2 = v_0 - \mu gt_0 ⇒ \mu = \frac{v_0}{2gt_0}\).