Question

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

Answer 1

If $H$ be the total height from where the body falls from rest, total time to fall is $t=\sqrt{\frac{2H}{g}}$
It is given that the body falls $\frac{H}{2}$ distance in time $(t-1)$, thus we have
$t-1=\frac{\sqrt{2\left(\frac{H}{2}\right)}}{g}$
Now, from above equations, eleminating $t$, we have
$\sqrt{\frac{2H}{g}}-\sqrt{\frac{H}{g}}=1$
or $\sqrt{\frac{H}{g}}(\sqrt{2}-1)=1$
or $\sqrt{\frac{H}{g}}=\frac{1}{0.414}$
or $H=58.34\text{m}$
Therefore, total time of fall is $t=\sqrt{\frac{2H}{g}}=\sqrt{\frac{116.68}{10}}=3.41\text{s}$

Aliter,
Taking downward direction as positive
For $AB$:
$s=ut+\frac{1}{2}a{t}^2$
or $\frac{H}{2}=\frac{1}{2}g(T-1{)}^2$ ... (1)
[Let the total time of fall be $T$]
For $AC:$
$s=ut+\frac{1}{2}a{t}^2$
or $H=\frac{1}{2}g{T}^2$ ... (2)
Dividing (2) and (1), we get
$2=\frac{T^2}{(T-1{)}^2}$
or $1.414=\frac{T}{T-1}$
or $1.414T-1.414=T$
or $0.414T=1.414$
or $T=3.41\text{s}$