Question

A body is dropped from a height of 2 metre .it penetreates into this sand on the ground through a distance of 10 M before coming to rest .what is the retardation of the body in sand.

Answer 1

Hello ,

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First calculate the velocity with which the stone penetrates the sand .

using v^2−u^2=2as


u = 0

s = 2 m

a = g = 10 m/s2

v^2−(0)^2 = 2×10× 2

v^2 = 40

$v = \sqrt{40} = 2 \sqrt{10} \: \: m \: {s}^{ - 1}$

Let the acceleration of the stone be constant and its value be a

Again applying v^2−u^2=2as

here,

s = 10 m

v = 0 m/s

$u = 2 \sqrt{10} \: m \: {s}^{ - 1}$

v^2−u^2=2as

0− 40 =2×a×10

a = -2 m/s2

The retardation of the stone will be -2 m/s2


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Hope it helps u !!!!

# Nikky