A hailstone falls from a height of #5# #km#. Ignoring air resistance, what is the hailstone's final velocity? The acceleration due to gravity is #9.8# #ms^-2#.

1 Answer
David G.
May 9, 2016

If we assume no air resistance the velocity is given by:

#v^2=u^2+2ad# and #u=0#, so, rearranging:

#v=sqrt(2ad)=sqrt(2*9.8*5000)=313# #ms^-1# or about #1127# #km^-1#

Explanation:

In order to answer this question it is necessary to assume that air resistance does not act. This is not a very realistic assumption.

Any real object falling has a 'terminal velocity' at which the air resistance acting upward balances the gravitational force acting downward. At that velocity, there is no further acceleration.

For a human skydiver this velocity is about #56# #ms^-1# or #200# #kmh^-1#. A hailstone is both smaller and lighter, so it is hard to guess its terminal velocity, but it would be less than #313# #ms^-1#.

Related questions
See all questions in Circular Motion
Impact of this question
3337 views around the world
You can reuse this answer
Creative Commons License