Question

A body is dropped from a height of 2m it penetrates into the sand on the ground through a distance of 10cm before coming to rest. What is the retardation of the body in the sand?
Answer must be 200m s^-2

Answer 1

Answer:

Explanation:

1/2 mv² = mgh

v = √2gh

v = √2*10*2

v = √2 * 3.14

v = 6.28 m/s

v² = u² + 2as

v = 0

u² = 2as

a = u² / 2s  = (2* 10 * 2 / 2*10) 100

= 200 m/s²

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