Question

A body falls from rest under gravity and travels half of its total path in the last second. Find the time of fall. (Take $g=10m{s}^{-2}$)

• A. $2.8s$
• B. $3.4s$
• C. $3.8s$
• D. $4.2s$

Answer 1

The correct option is B $3.4s$

Let $H$ be the total height through which the body falls .
Time taken by the body to fall through height $H$
$t=\sqrt{\frac{2H}{g}}$ ...(i)
It is given that the body falls $H/2$ distance in time $(t-1)$. Thus we have, $t-1=\sqrt{\frac{2\left(\frac{H}{2}\right)}{g}}$..(ii)
Using (i) and (ii), we have
$\sqrt{\frac{2H}{g}}-\sqrt{\frac{H}{g}}=1$
$⟹\sqrt{\frac{H}{g}}(\sqrt{2}-1)=1$
$⟹\sqrt{\frac{H}{g}}=\frac{1}{0.414}$
Given $H=58m$
$\therefore$ Time of fall is $t=\sqrt{\frac{2H}{g}}=\frac{1.414}{0.414}=3.4s$

Alternative solution :


Taking downward as positive and total time of fall be $T$.
For AB:
$s=ut+\frac{1}{2}a{t}^2$
$\begin{array}{}\frac{H}{2}=\frac{1}{2}g(T-1{)}^2& \text{(1)}\end{array}$
For AC:
$s=ut+\frac{1}{2}a{t}^2$
$\begin{array}{}H=\frac{1}{2}g{T}^2& \text{(2)}\end{array}$
Dividing equation $\left(2\right)$ by $\left(1\right)$, we get
$2=\frac{T^2}{(T-1{)}^2}$
$1.414=\frac{T}{T-1}$
$0.414T=1.414$
$\Rightarrow T=3.4s$