Question

A body is freely falling under the action of gravity. It covers half the total distance in the last second of its fall. If it falls for $n$ second, then the value of $n$ is

• A. $\sqrt{2}$
• B. $2+\sqrt{2}$
• C. $2$
• D. $2-\sqrt{2}$

Answer 1

The correct option is B $2+\sqrt{2}$

Let $h$ be the total distance covered in $n$ second then,
for free fall $h=\frac{1}{2}g{n}^2$
As per the question, distance covered by it in last second i.e. $(n-1{)}^{th}\text{sec}$ is given by
$h^{{}^{\prime }}=\frac{h}{2}=\frac{\frac{1}{2}g{n}^2}{2}=\frac{1}{4}g{n}^2$
Thus, we have
$\frac{1}{2}g(n-1{)}^2=\frac{1}{4}g{n}^2$
Solving above equation we get a quadratic equation in $n$ as
$n^2-4n+2=0$
$\Rightarrow n=\frac{-(-4)\pm \sqrt{(-4{)}^2-4\times 2}}{2}$
$\Rightarrow n=\frac{4\pm 2\sqrt{2}}{2}=2\pm \sqrt{2}$
Now, $n-1>0$ if $n=2-\sqrt{2}$
$\Rightarrow (2-\sqrt{2})-1>0\Rightarrow 1-\sqrt{2}<0$
Hence, the value of $n$ is $2+\sqrt{2}$.