Answer

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**Hint:**We can use position time relation which is given by

$ S=ut+\dfrac{1}{2}a{{t}^{2}} $

Here, S is distance travelled in time t,

u is the initial velocity,

a is the acceleration and t is time taken.

In case of free fall, if downward direction is taken positive and object is released from rest, then

$ u=0,\text{ }a=g=10m/{{s}^{2}},\text{ }S=h $

In case of free fall, if downward direction is taken negative and object is released from set, then

$ u=0,\text{ }a=-g=-10m/{{s}^{2}},\text{ }S=-h $ .

**Complete step by step solution**

We have, given

Stone is falling freely under gravity.

It means, if we will take downward as negative

Then, $ a=-g $

$ S=-{{h}_{1}} $

Case I: Where $ {{h}_{1}} $ is the distance covered in first seconds,

Here, stone is falling freely hence take initial velocity as zero.

$ u=0 $ , t=5s is given

Then use distance/position-time relation

$ S=ut+\dfrac{1}{2}a{{t}^{2}} $

Here, $ a=-g $ , means acceleration due to gravity.

Put all the above values:

$ {{h}_{1}}=\dfrac{25}{2}g $ ; $ -{{h}_{1}}=0\left( 5 \right)+\dfrac{1}{2}\left( -g \right){{\left( 5 \right)}^{2}} $

$ -{{h}_{1}}=-\dfrac{1}{2}\times g{{\left( 25 \right)}^{{}}} $

$ {{h}_{1}}=\dfrac{25}{2}g $ ----------(2)

Now in Case II:

Distance travelled by stone in next 5 seconds,

Now total distance covered will be, $ S=-\left( {{h}_{1}}+{{h}_{2}} \right) $ [negative sign shows just direction of measurement]

And total time taken, t=5+5=10s

Put the above values in eq, (1)

$ -\left( {{h}_{1}}+{{h}_{2}} \right)=0\left( 10 \right)+\dfrac{1}{2}\left( -g \right){{\left( 10 \right)}^{2}} $

$ \left( {{h}_{1}}+{{h}_{2}} \right)=\dfrac{1}{2}g\left( 100 \right)=\dfrac{100}{2}g $ ---------(3)

Case III:

Distance travelled by stone in next 5 seconds,

Total distance covered $ S=-\left( {{h}_{1}}+{{h}_{2}}+{{h}_{3}} \right) $

Total time taken $ t=5+5+5=15s $

Put the values in eq. (1)

$ -\left( {{h}_{1}}+{{h}_{2}}+{{h}_{3}} \right)=-\dfrac{1}{2}g{{\left( 15 \right)}^{2}} $

$ \left( {{h}_{1}}+{{h}_{2}}+{{h}_{3}} \right)=\dfrac{225}{2}g $ ------- (4)

Solve eq. (2) and (3)

$ \left( {{h}_{1}}+{{h}_{2}} \right)-{{h}_{1}}=\dfrac{100}{2}g-\dfrac{25}{2}g $

$ {{h}_{2}}=\dfrac{75}{2}g $ -------- (5)

Solve eq. (3) and (4),

$ \left( {{h}_{1}}+{{h}_{2}}+{{h}_{3}} \right)-\left( {{h}_{1}}+{{h}_{2}} \right)=\dfrac{225}{2}g-\dfrac{100}{2}g $

$ {{h}_{3}}=\dfrac{125}{2}g $ --------- (6)

Write eq. (5) and (6) in terms of

$ {{h}_{2}}=\dfrac{25}{2}\times 3g\to {{h}_{2}}=3{{h}_{1}} $ ---------- (7)

$ {{h}_{3}}=\dfrac{25}{2}\times 5g\to {{h}_{3}}=5{{h}_{1}} $ ----------- (8)

From eq. (7) and (8),

$ {{h}_{1}}={{\dfrac{{{h}_{2}}}{3}}_{{}}}=\dfrac{{{h}_{3}}}{5} $ This is required result.

**Therefore option (B) is the correct answer.**

**Note**

We can also, use direct formula, Distance travelled in the second by,

$ {{D}_{n}}=u+\dfrac{a}{2}\left( 2n-1 \right) $

Here, n is nth second, above question n is given as 5s.

Using the above formula we will get the same answer.

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