A stone is thrown from the top of a tower of height 100m.The stone penetrates in the sand on the ground through a distance of 2m.Calculate the retardtion of the stone.

As the stone is dropped from a height of 100 m, its potential energy is completely converted into kinetic energy just before hitting the ground.

So, just before hitting the ground the kinetic energy is, ½ mu² = mgh

=> u² = (2gh)

=> u² = (2 × 9.8 × 100) = 1960

Now, the stone comes to rest after penetrating 2 m. Let the retardation be ‘a’.

So,

v² = u² – 2as

=> 0 = 1960 – 2(a)(2)

=> a = 1960/4 = 490 m/s²

The direction of this retardation is opposite to the direction of motion of the stone.