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**Hint:**When a body falls from the rest its initial velocity is zero and as the body is falling freely therefore the acceleration acting on the body is acceleration due to gravity. The velocity of the body will keep on increasing as the acceleration acting is constant and therefore the distance covered by the body will also keep on increasing.

**Formula used:**The formula of the second relation of the Newton’s law of motion is given by,

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

Where displacement is s the initial velocity is u the total time taken is t and the acceleration is a.

The formula for the ${S_n}$ distance travelled in ${n^{th}}$ second is given by,

$ \Rightarrow {S_n} = u + \dfrac{a}{2}\left[ {2n - 1} \right]$

Where ${S_n}$ is the total distance travelled the initial velocity is u the acceleration on the body is a and n is the ${n^{th}}$ second.

**Complete step by step solution:**

It is given in the problem that a body falls from rest and in the last second of its fall it covers half of the total distance then we need to find the total time of the motion.

The formula for the ${S_n}$ distance travelled in ${n^{th}}$ second is given by,

$ \Rightarrow {S_n} = u + \dfrac{a}{2}\left[ {2n - 1} \right]$

Where ${S_n}$ is the total distance travelled the initial velocity is u the acceleration on the body is a and n is the ${n^{th}}$ second.

As the distance covered is half of the total distance covered in the last second and the acceleration is acceleration due to gravity therefore we get,

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

$ \Rightarrow \dfrac{h}{2} = \dfrac{g}{2}\left( {2n - 1} \right)$

$ \Rightarrow h = g\left( {2n - 1} \right)$

$ \Rightarrow h = 10 \times \left( {2t - 1} \right)$………eq. (1)

The formula of the second relation of the Newton’s law of motion is given by,

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

Where displacement is s the initial velocity is u the total time taken is t and the acceleration is a.

As the body is initially at rest and the acceleration due to gravity is $g = 10\dfrac{m}{{{s^2}}}$ therefore we get,

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

$ \Rightarrow h = \dfrac{1}{2}\left( {10} \right){\left( t \right)^2}$………eq. (2)

Equating equation (1) and equation (2) we get.

$ \Rightarrow \dfrac{1}{2}{t^2} = \left( {2t - 1} \right)$

$ \Rightarrow {t^2} = 2\left( {2t - 1} \right)$

$ \Rightarrow {t^2} = 4t - 2$

$ \Rightarrow {t^2} - 4t + 2 = 0$

The formula for solving quadratic rule is given by,

$ \Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

Therefore solving the equation we get,

$ \Rightarrow {t^2} - 4t + 2 = 0$

Here $a = 1$, $b = - 4$ and $c = 2$.

$ \Rightarrow t = \dfrac{{ - \left( { - 4} \right) \pm \sqrt {{{\left( { - 4} \right)}^2} - 4 \cdot \left( 1 \right) \cdot \left( 2 \right)} }}{{2 \cdot \left( 1 \right)}}$

$ \Rightarrow t = \dfrac{{4 \pm \sqrt {16 - 8} }}{2}$

$ \Rightarrow t = \dfrac{{4 \pm \sqrt 8 }}{2}$

$ \Rightarrow t = \dfrac{{4 \pm 2\sqrt 2 }}{2}$

$ \Rightarrow t = 2 \pm \sqrt 2 $.

The total time taken is equal to $t = 2 \pm \sqrt 2 $s.

**The correct answer for this problem is option (D).**

**Note:**It is advisable to students to understand and remember the formulas of Newton’s law motion as it is helpful in solving problems of these kinds. It is given in the problem that the body covers half of the total distance in the last second of the fall of the body.

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