Question

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

Answer 1

Correct Question

A body is dropped from a height of 2 metre it penetrates into the sand on the ground through a distance 10 m or 10 cm before coming to rest what is the retardation of the body in sand.

Solution -

A body is dropped from a height of 2 metre it penetrates into the sand on the ground through a distance 10 before coming to rest.

We have to find the retardation of the body in sand.

From above data we have; height (h) = 2 m, initial velocity (u) of the body is 0 m/s and acceleration due to gravity (a) is 10 m/s².

Using the Third Equation Of Motion:

v² - u² = 2as

v² - (0)² = 2(10)(2)

v² = 40

In second part it's given that, The body penetrates into the sand on the ground through a distance 10 before coming to rest.

Here; distance or height (s) is 10 m, initial velocity becomes final velocity i.e. √40 m/s and final velocity is 0 m/s.

Again, using the Third Equation Of Motion:

v² - u² = 2as

Substitute the known values,

(0)² - 40 = 2(a)(10)

-40 = 20a

-2 = a

(Negative sign retardation)

Therefore, the retardation of the body in sand is 2 m/s².

If the value of distance was in cm then s = 0.1 m.

So,

(0)² - 40 = 0.2a

-40 = 0.2a

-200 = a

Therefore, the value of retardation of the body in sand is 200 m/s².

$\rule{200}1$

Note:- If we take value of a = 9.8 m/s² then

If s = 10 m then a = 1.96 m/s² and if s = 0.1 m then a = 196 m/s².