A body falls freely from rest. It covers as much distance in the last second of its motion as covered in the first three seconds. Which of the following is the time taken by the body to fall off?

A body falls freely from rest. It covers as much distance in the last second of its motion as covered in the first three seconds. The body has fallen for a time of $5\mathit{ s}.$
When a ball falls downward, a gravitational force acts on it which supports its motion in the downward direction. So, acceleration due to gravity acts on the ball in the direction of its motion. 
According to the question, $\mathrm{displacement} \mathrm{in} \mathrm{first} 3 \mathrm{seconds}=\mathrm{displacement} \mathrm{in} \mathrm{last} \mathrm{second}$
Here, initial velocity, $u=0\mathit{ m}/s$
Acceleration, $a=g$ (as the body is falling freely)
$\Rightarrow a= 10\mathit{ m}/s^2$ 

Distance travelled in $3\mathit{ s}$, 
     $\mathrm{s}=\mathrm{ut}+\mathrm{ }\frac{1}{2}{\mathrm{at}}^2$
$\Rightarrow s=0 \times 3+ \frac{1}{2} \times 10 \times 3 \times 3$
$\Rightarrow s=5 \times 9$
$\Rightarrow s=45\mathit{ m}$
So, displacement in last second $=45\mathit{ m}$
Velocity in last second $=g(t-1)$
Distance covered in last second,
  $\mathrm{s}=\mathrm{ut}+\mathrm{ }\frac{1}{2}{\mathrm{at}}^2$
$\Rightarrow \mathrm{s}=\mathrm{g}(\mathrm{t}-1)\mathrm{ }\times \mathrm{t}+\mathrm{ }\frac{1}{2}\mathrm{ }\times \mathrm{a} \times \mathrm{t} \times \mathrm{t}$
$\Rightarrow 45=10 \left(t-1\right) \times 1+ \frac{1}{2} \times 10 \times 1 \times 1$
$\Rightarrow 45=10 \left(t-1\right)+5$
$\Rightarrow 10\left(t-1\right)=40$
$\Rightarrow \mathrm{t}-1=\mathrm{ }\frac{40}{10}$
$\Rightarrow \mathrm{t}-1=4$
$\Rightarrow \mathrm{t}=5 \mathrm{s}$