Height/velocity

Question

hail stones are falling from a height of 500 meters. each hailstone has a mass of 30 grams. how much kinetic energy, in joules, does one of these hail stones have when it hits the ground?  what velocity, in m/s, does the hailstone have?

Answer 1

We are, presumably, neglecting air resistance in this problem.

 

The core of this problem appears to be one of energy conservation.  The hailstone when 'dropped' from 500 meters has a certain amount of gravitational potential energy, which is gradually turned into kinetic energy as it falls, with the total energy being preserved.  The gravitational potential of a mass m at a height h is given by PE = mgh, (g being acceleration due to gravity, 9.8 m/s²).

 

For one of our hailstones, m is 30 grams = .03 kg, and h = 500 m, so PE = (.03)(9.8)(500) = 147 joules.

 

When it hits the ground, the height is zero, so the gravitational potential energy is also zero - all 147 joules of potential energy that it initially had have been turned into kinetic energy.  Hence, at this point, its kinetic energy is 147 joules.

 

We can figure out what velocity this kinetic energy implies by reference to the kinetic energy equation, KE = .5mv².

m is again .03 kg, and KE, as just stated, is 147 J.  Hence,

 

147 = .5*.03*v^2

v^2 = 147/(.015) = 9800

v = √9800 ~= 99 m/s.

 

(Note that this is the same result you get when applying a constant acceleration of 9.8 m/s² over a distance of 500 meters.  That's governed by the equation v² = v₀² + 2as:

v² = 0² + 2(9.8)(500) = 9800,

v ~= 99 m/s)