Question

A body travels half of its total path in the last second of its fall from rest. If $t$ and $H$ are the time and height of its fall (take $g=10{\text{m/s}}^2$). Then,

• A. $t=2.4\text{s}$
• B. $t=3.4\text{s}$
• C. $H=28.8\text{m}$
• D. $H=58.3\text{m}$

Answer 1

Answer: (B), (D)

$H=58.3\text{m}$

If $H$ be the total height by which the body falls, then the total time to fall is given by
$H=ut+\frac{1}{2}g{t}^2$
As $u=0$
$\Rightarrow t=\sqrt{\frac{2H}{g}}.....\text{(i)}$
It is given that the body falls $\frac{H}{2}$ distance in time $(t-1)$, thus we have
$t-1=\sqrt{\frac{2\left(\frac{H}{2}\right)}{g}}......\text{(ii)}$
Subtracting $\text{(i)}$ from $\text{(ii)}$, we get
$\sqrt{\frac{2H}{g}}-\sqrt{\frac{2\left(\frac{H}{2}\right)}{g}}=1$
Or $\sqrt{\frac{H}{g}}(\sqrt{2}-1)=1$
Or $\sqrt{\frac{H}{g}}=\frac{1}{0.414}$
Or $H=58.3\text{m}$
Total time of fall is $t=\sqrt{\frac{2H}{g}}=3.4\text{s}$