Answer
The skater's initial velocity was 12 m/s.
Work Step by Step
The force of kinetic friction opposed the motion of the skater and brought the skater to a stop. Let's find the magnitude of acceleration:
$ma = F_f$
$ma = mg ~\mu_k$
$a = g ~\mu_k$
$a = (9.8 ~m/s^2)(0.10)$
$a = 0.98 ~m/s^2$
Since the force of kinetic friction opposed the skaters motion, the acceleration is $-0.98 ~m/s^2$. Therefore,
$v^2 - v_0^2 = 2ax$
$-v_0^2 = (2)(-0.98 ~m/s^2)(75 ~m)$
$v_0 = \sqrt{(2)(0.98 ~m/s^2)(75 ~m)}$
$v_0 = 12 ~m/s$
The skater's initial velocity was 12 m/s.