Kinematics with Friction

Let $a$ be the deceleration due to the frictional force of the horizontal surface. Then the displacement of the object time $t \text{ s}$ after being launched is given by:

$\begin{aligned} s(t) & = v_0t - \frac 12 at^2 \\ \implies s(4) & = 20(4) - \frac 12 (16)a \\ s(5) & = 20(5) - \frac 12 (25)a \\ \implies s(5) - s(4) & = 20 - \frac 92 a \\ 5 & = 20 - \frac 92 a \\ \implies a & = \frac {10}3 \text{ m/s}^2 \end{aligned}$

Let the mass of the object be $m \text{ kg}$ and the coefficient of the surface be $\mu$ , then we have:

$\begin{aligned} \mu mg & = ma \\ \implies \mu & = \frac ag = \frac {10}3 \times \frac 1{10} = \boxed{\dfrac 13} \end{aligned}$