Question

A freely falling body takes $t$ second to travel first $(\frac{1}{x}{)}^{th}$ distance. Then, time of descent is

• A. $\frac{t}{\sqrt{x}}$
• B. $t\sqrt{x}$
• C. $\frac{\sqrt{x}}{t}$
• D. $\frac{1}{t\sqrt{x}}$

Answer 1

Answer: (B)

$t\sqrt{x}$

Let total distance be s
So, $s\left(\frac{1}{x}\right)=\frac{1}{2}g{t}^2$
$s=\frac{x}{2}g{t}^2$
Now, time of descent, $T=\sqrt{\frac{2s}{g}}$
$T=\sqrt{\left(\frac{2}{g}\right)\times \frac{x}{2}g{t}^2}$
$T=t\sqrt{x}$