Question

$A$ , $B$ , $C$ are points on a vertical line such that $AB \, = \, BC$ . If a body is dropped from rest at $A$ , and $t_{1}$ and $t_{2}$ are the time to travel for distance $AB$ and $BC$ , then ratio $\left(\frac{t_{2}}{t_{1}}\right)$ is

• A. $\sqrt{2}+1$
• B. $\sqrt{2}-1$
• C. $2\sqrt{2}$
• D. $\frac{1}{\sqrt{2} + 1}$

Answer 1

Answer: (B)

$-h=0-\frac{1}{2}gt^{2}$
$t=\sqrt{\frac{2 h}{g}}\ldots \ldots ..\left(1\right)$
$-\frac{h}{2}=0-\frac{1}{2}gt_{1}^{2}$
$t_{1}=\sqrt{\frac{h}{g}}\ldots \ldots ..\left(2\right)$
$so t_{2}=t-t_{1}$
$=\sqrt{\frac{2 h}{g}}-\sqrt{\frac{h}{g}}$
$t_{2}=\sqrt{\frac{h}{g}}\left(\sqrt{2} - 1\right)\ldots \ldots .\left(3\right)$
equation $3/1$
$\frac{t_{2}}{t_{1}}=\frac{\sqrt{\frac{h}{g}} \left(\sqrt{2} - 1\right)}{\sqrt{\frac{h}{g}}}$
$\frac{t_{2}}{t_{1}}=\sqrt{2}-1$