Answer:6.92 m/s
Explanation:
Given
initial velocity [tex]u=16.1 m/s[/tex]
coefficient of kinetic Friction [tex]\mu _k=0.72[/tex]
time interval [tex]t=1.3 s[/tex]
Let final velocity be [tex]v m/s[/tex]
maximum acceleration[tex]=-\mu _k\times g[/tex]
[tex]a=-0.72\times 9.8\approx 7.056 m/s^2[/tex]
using [tex]v=u+at[/tex]
[tex]v=16.1-7.056\times 1.3[/tex]
[tex]v=16.1-9.17[/tex]
[tex]v=6.92 m/s[/tex]