Answer:
a) 182.54m/s
b) In that case it would not be safe to walk outside during a rainstorm.
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.
[tex]v^2=u^2+2gh.................(1)[/tex]
were v is the velocity with which they strike the ground, g is acceleration due to gravity taken as [tex]9.8m/s^2[/tex], h is the height from which the drops fall.
Given;
[tex]h=1700m\\[/tex], u = 0m/s since the raindrops are assumed to drop freely from rest.
Therefore;
[tex]v^2=0^2+2*9.8*1700\\v^2=33320\\v=\sqrt{33320}\\ v=182.54m/s[/tex]
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.
