Answer:
a) 182.54m/s
b) In that case it would not be safe to walk outside during a rainstorm.
Step-by-step explanation:
We use the third equation of motion of free fall under gravity to determine the velocity with which the raindrops will strike the ground.

were v is the velocity with which they strike the ground, g is acceleration due to gravity taken as
, h is the height from which the drops fall.
Given;
, u = 0m/s since the raindrops are assumed to drop freely from rest.
Therefore;

At this velocity I think it will not be safe to walk outside during a rainstorm because millions of water molecules falling at this speed will cumulatively posses a very high amount of kinetic energy that will impact whoever comes under it. The water molecules may also appear as a fast flowing river.
It is only if the mass of water molecules is completely neglected that it might still be safe to walk under a rainstorm in such case.