Question

A 5kg stone falls from a height 100m and penetrates 2m in a layer of sand .the time of penetration is:

Answer 1

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Answer 2

Answer:

The time of penetration is 0.09 sec.

Explanation:

Given that,

mass of stone = 5 kg

Height h = 100 m

Distance of penetrates = 2 m

We need to calculate the initial velocity

When the stone reaches the ground then its potential energy is converted into kinetic energy.

Potential energy = kinetic energy

$mgh=\dfrac{1}{2}mu^2$

$u=\sqrt{2gh}$

When the stone penetrate in a layer of sand then the stone experiences deceleration a, stop the stone after 2 m

So, the final velocity of the stone is zero

Using equation of motion

$v^2=u^2+2as$

$0=(\sqrt{2gh})^2+2as$

$a=-\dfrac{gh}{s}$

The time of penetration is

Using equation of motion again

$v =u-at$

$t =\dfrac{v-u}{a}$

$t=\dfrac{0-\sqrt{(2gh)}}{\dfrac{-gh}{s}}$

$t = \dfrac{s\sqrt{2}}{\sqrt{gh}}$

Put the value of g,s and h in to the formula

$t =\dfrac{2\sqrt{2}}{\sqrt{9.8\times100}}$

$t =0.09\ s$

Hence, The time of penetration is 0.09 sec.