Question

A body dropped freely has covered half of the total distance in the last second. The total journey time is

• A. $(2+\sqrt{2})s$
• B. $(2-\sqrt{2})s$
• C. $2s$
• D. $(2+\sqrt{3})s$

Answer 1

Answer: (A)

$(2+\sqrt{2})s$

Let total height $=h,$ let total time $=n$

$\frac{h}{2}=\left[u\right(n)+\frac{1}{2}g{n}^2]-\left[u\right(n-1)+\frac{1}{2}g(n-1{)}^2]$

$\frac{h}{2}=\frac{1}{2}g(2n-1)$

$h=g(2n-1)......\left(1\right)$

Also, $h=\frac{1}{2}g{n}^2.....\left(2\right)$

Equating $\left(1\right)$ and $\left(2\right)$,

$g\frac{n^2}{2}=g(2n-1)$

$n^2-4n+2=0$

$(n-2{)}^2=2\Rightarrow n=\pm \sqrt{2}+2$

$n=2+\sqrt{2},2-\sqrt{2}$

Since $n>1\Rightarrow n=2+\sqrt{2}$