Question

In the last second of freefall a body covers 3/4 of its total path .Then the height from which that body is released​

Answer 1

Let the total time in free fall be $t$

Then, the distance covered just before the last second would be $\frac{1}{4}th$ the total distance, or $t-1$ seconds.

$→ h(t) = \frac{1}{{2gt}^{2}}$

$→ h(t-1) = \frac{1}{{2g(t-1)}^{2}}$

$→ \frac{1}{4} \frac{1}{{2gt}^{2}} = \frac{1}{{2 g(t-1)}^{2}}$

$→ \frac{g}{{2t}^{2}}=2g({t}^{2} -2t+1)$

$→ {t}^{2} = {4t}^{2} -8t+4$

$→ {3t}^{2} -8t+4=0$

$→ (3t-2)(t-2)=0$

$→ t= \frac{2}{3} or 2$

We can ignore the $\frac{2}{3}rds$ seconds, and get a value for $t$ as $2$ seconds. So, the body fell for $2$ seconds, and covered $4.9(4)$ or $19.6m$