# An object is dropped from a height of 20m above the ground what will be its speed just before it hits the ground?

The method that we using here is of energy conservation.

Let us assume the mass of the object as m kg

Since the object is placed at height of 20 m above the ground, it possess potential energy . It is given by

**PE = m x g x h = 20 mg **

Let us assume that it has a velocity v just before striking the ground.

Then the kinetic energy of the object be, ** KE= ½ m v²**

Applying conservation of energy principle we can write,

Potential Energy of the object = Kinetic Energy of the object

20 m g = ½ m v²

40 m g = m v²

v² = 40 g.

Taking g = 10 m/s²

v² = 400

And hence

v = ±20 m/s

The speed with which the object would strike the ground is 20 m/s directed in downward direction.