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#.

Answer 1

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#

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#.

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Axel Acevedo

The final velocity of the hailstone is 98 m/s.

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.