We can solve this problem using the law of conservation of momentum, which states that the total momentum of a system remains constant if there are no external forces acting on it.
The initial momentum of the system is:
p = m1v1 + m2v2
where m1 is the mass of the archer, v1 is the initial velocity of the archer (before shooting), m2 is the mass of the arrow, and v2 is the initial velocity of the arrow (after being shot).
p = (90.0 kg)(0 m/s) + (0.12 kg)(73 m/s)
p = 8.76 kg m/s
According to the law of conservation of momentum, the final momentum of the system (after the arrow is shot) must also be 8.76 kg m/s.
Let's assume that the recoil velocity of the archer is v'. The final momentum of the system is then:
p' = (90.0 kg)(v') + (0.12 kg)(0 m/s)
p' = 90v' kg m/s
Setting p' equal to the initial momentum, we have:
p' = p
90v' kg m/s = 8.76 kg m/s
Solving for v', we get:
v' = 0.09733 m/s
Therefore, the recoil velocity of the archer is approximately 0.097 m/s.