Question

A body starts falling from height h and travels distance h/2 during last second of motion,then time of travel(in sec)is__________.

Answer 1

time = t
height = h
h = 1/2 gt²
2h = gt²

time to travel the first half = (t-1)
so, 
h/2  = 1/2 g (t-1)²
h = g (t-1)²

divide both equations
2= t² / (t-1)²

taking under rot on both sides
t/t-1 = √2 = 1.414
t = 1.414 t - 1.414
t = 3.41 s

Answer 2

$\textbf{ Hello Mate! }$

Let the height be "h" as given and let the time be "t" during whole journey.

As the object is dropped from certain height to "u" = 0

By second equation of motion we get,

$s = ut + \frac{1}{2} a {t}^{2} \\ h = (0)(t) + \frac{1}{2} (10)( {t)}^{2} \\ 2h = g {t}^{2} ......(1)$

Now for last second, distance was h / 2 and u = 0 and time was t - 1

$\frac{h}{2} = (0)(t) + \frac{1}{2} g {(t - 1)}^{2} \\ h = g {(t - 1)}^{2} ........(2)$
By dividing (1) by (2)

$\frac{2h}{ h } = g {t}^{2} \times \frac{1}{g {(t - 1)}^{2} } \\ 2 = {( \frac{t}{t - 1}) }^{2}$

On doing under root of equation we get

$1.414 = \frac{t}{t - 1} \\ 1.414t - 2 = t \\ 1.414t - t = 1.414 \\ 0.414t = 1.414 \\ t = 3.42\: seconds$

$\boxed{ \textsf{ \red{ =>Time=3.42\:seconds}}}$

Have great future ahead!